100 results found for "distinct-real-roots" in Class 10.
\(x^2+12x+\lambda=0\) की जड़ें वास्तविक भिन्न और दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?
For \(x^2+12x+\lambda=0\) to have real distinct roots and both negative roots, what is the correct condition on \(\lambda\)?
#quadratic-roots
#negative-distinct-roots
#condition
A \(0<\lambda<36\)
B \(\lambda=36\)
C \(\lambda>36\)
D \(\lambda<0\)
Explanation opens after your attempt
Correct Answer
A. \(0<\lambda<36\)
Step 1
Concept
For both roots to be negative, the sum (-12) and product \(\lambda>0\) are needed. For real distinct roots, \(144-4\lambda>0\), so \(0<\lambda<36\).
Step 2
Why this answer is correct
The correct answer is A. \(0<\lambda<36\). For both roots to be negative, the sum (-12) and product \(\lambda>0\) are needed. For real distinct roots, \(144-4\lambda>0\), so \(0<\lambda<36\).
Step 3
Exam Tip
दोनों ऋणात्मक जड़ों के लिए योग (-12) और गुणनफल \(\lambda>0\) चाहिए। वास्तविक भिन्न जड़ों के लिए \(144-4\lambda>0\), इसलिए \(0<\lambda<36\)।
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\(x^2+10x+\lambda=0\) की जड़ें वास्तविक भिन्न और दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?
For \(x^2+10x+\lambda=0\) to have real distinct roots and both negative roots, what is the correct condition on \(\lambda\)?
#quadratic-roots
#negative-distinct-roots
#condition
A \(\lambda<0\)
B \(0<\lambda<25\)
C \(\lambda=25\)
D \(\lambda>25\)
Explanation opens after your attempt
Correct Answer
B. \(0<\lambda<25\)
Step 1
Concept
For both roots to be negative, the sum (-10) and product \(\lambda>0\) are needed. For real distinct roots, \(100-4\lambda>0\), hence \(0<\lambda<25\).
Step 2
Why this answer is correct
The correct answer is B. \(0<\lambda<25\). For both roots to be negative, the sum (-10) and product \(\lambda>0\) are needed. For real distinct roots, \(100-4\lambda>0\), hence \(0<\lambda<25\).
Step 3
Exam Tip
दोनों ऋणात्मक जड़ों के लिए योग (-10) और गुणनफल \(\lambda>0\) चाहिए। वास्तविक भिन्न जड़ों के लिए \(100-4\lambda>0\), इसलिए \(0<\lambda<25\)।
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\(x^2+2x+\lambda=0\) की जड़ें वास्तविक और भिन्न हों तथा दोनों ऋणात्मक हों, तो \(\lambda\) पर सही शर्त क्या है?
For \(x^2+2x+\lambda=0\) to have real distinct roots and both negative roots, what is the correct condition on \(\lambda\)?
#quadratic-roots
#negative-distinct-roots
#condition
A \(0<\lambda<1\)
B \(\lambda>1\)
C \(\lambda<0\)
D \(\lambda=1\)
Explanation opens after your attempt
Correct Answer
A. \(0<\lambda<1\)
Step 1
Concept
For both roots to be negative, the sum (-2) and product \(\lambda>0\) are needed. For real distinct roots, \(4-4\lambda>0\), hence \(0<\lambda<1\).
Step 2
Why this answer is correct
The correct answer is A. \(0<\lambda<1\). For both roots to be negative, the sum (-2) and product \(\lambda>0\) are needed. For real distinct roots, \(4-4\lambda>0\), hence \(0<\lambda<1\).
Step 3
Exam Tip
दोनों ऋणात्मक जड़ों के लिए योग (-2) और गुणनफल \(\lambda>0\) चाहिए। वास्तविक भिन्न जड़ों के लिए \(4-4\lambda>0\), इसलिए \(0<\lambda<1\)।
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(x-2 -2(k+1)x+k-2 =0) की जड़ें वास्तविक और भिन्न हों, तो (k) पर सही शर्त क्या है?
For (x-2 -2(k+1)x+k-2 =0) to have real and distinct roots, what is the correct condition on (k)?
#quadratic-roots
#distinct-roots
#discriminant
A \(k>-\frac{1}{2}\)
B \(k\ge-\frac{1}{2}\)
C \(k<-\frac{1}{2}\)
D \(k\le-\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(k>-\frac{1}{2}\)
Step 1
Concept
For real and distinct roots, (D>0) is needed. Here (D=4(2k+1)), so \(k>-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(k>-\frac{1}{2}\). For real and distinct roots, (D>0) is needed. Here (D=4(2k+1)), so \(k>-\frac{1}{2}\).
Step 3
Exam Tip
वास्तविक और भिन्न जड़ों के लिए (D>0) चाहिए। यहाँ (D=4(2k+1)), इसलिए \(k>-\frac{1}{2}\)।
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समीकरण (x-2 -2(m-4 )x+m-2 -16=0) के मूल वास्तविक और भिन्न कब होंगे?
When will the roots of (x-2 -2(m-4 )x+m-2 -16=0) be real and distinct?
#quadratic equations
#parameter
#real distinct roots
A (m<4)
B (m>4)
C (m=4)
D \(m\le -4\)
Explanation opens after your attempt
Step 1
Concept
Here (D=32(4-m)). For real and distinct roots (D>0), so (m<4).
Step 2
Why this answer is correct
The correct answer is A. (m<4). Here (D=32(4-m)). For real and distinct roots (D>0), so (m<4).
Step 3
Exam Tip
यहाँ (D=32(4-m)) है। वास्तविक और भिन्न मूलों के लिए (D>0), इसलिए (m<4)।
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निम्न में से किस समीकरण के दो वास्तविक और असमान मूल हैं?
Which of the following equations has two real and distinct roots?
#quadratic-equations
#choose-equation
#real-distinct-roots
A \(x^2-11x+18=0\)
B \(x^2+4x+4=0\)
C \(x^2+2x+6=0\)
D \(x^2-8x+16=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-11x+18=0\)
Step 1
Concept
In option (A), (D=(-11)2 -4(1)(18)=49). When (D>0), two distinct real roots exist.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-11x+18=0\). In option (A), (D=(-11)2 -4(1)(18)=49). When (D>0), two distinct real roots exist.
Step 3
Exam Tip
विकल्प (A) में (D=(-11)2 -4(1)(18)=49) है। (D>0) होने पर दो असमान वास्तविक मूल मिलते हैं।
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समीकरण \(x^2-2mx+3m=0\) के वास्तविक और भिन्न मूलों के लिए (m) पर क्या शर्त है?
What condition on (m) gives real and distinct roots for \(x^2-2mx+3m=0\)?
#quadratic equations
#real distinct roots
#interval
A (m<0) या (m>3) / (m<0) or (m>3)
B (0<m<3)
C (m=0) या (m=3) / (m=0) or (m=3)
D (m>0)
Explanation opens after your attempt
Correct Answer
A. (m<0) या (m>3) / (m<0) or (m>3)
Step 1
Concept
Here (D=4m(m-3 )). From (D>0), (m<0) or (m>3).
Step 2
Why this answer is correct
The correct answer is A. (m<0) या (m>3) / (m<0) or (m>3). Here (D=4m(m-3 )). From (D>0), (m<0) or (m>3).
Step 3
Exam Tip
यहाँ (D=4m(m-3 )) है। (D>0) से (m<0) या (m>3) मिलता है।
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यदि (x-2 -2\(\alpha+2\)x+\alpha-2 =0) के मूल वास्तविक और भिन्न हैं, तो \(\alpha\) पर शर्त क्या है?
If (x-2 -2\(\alpha+2\)x+\alpha-2 =0) has real and distinct roots, what is the condition on \(\alpha\)?
#quadratic equations
#alpha
#real distinct roots
A \(\alpha>-1\)
B \(\alpha<-1\)
C \(\alpha=-1\)
D \(\alpha>2\) केवल / \(\alpha>2\) only
Explanation opens after your attempt
Correct Answer
A. \(\alpha>-1\)
Step 1
Concept
(D=4\(\alpha+2\)2 -4\alpha-2 =16\(\alpha+1\)). From (D>0), \(\alpha>-1\).
Step 2
Why this answer is correct
The correct answer is A. \(\alpha>-1\). (D=4\(\alpha+2\)2 -4\alpha-2 =16\(\alpha+1\)). From (D>0), \(\alpha>-1\).
Step 3
Exam Tip
(D=4\(\alpha+2\)2 -4\alpha-2 =16\(\alpha+1\)) है। (D>0) से \(\alpha>-1\) मिलता है।
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किस शर्त पर \(x^2-2sx+s+2=0\) के मूल वास्तविक और भिन्न होंगे?
Under which condition will \(x^2-2sx+s+2=0\) have real and distinct roots?
#quadratic equations
#interval condition
#real distinct roots
A (s<-1) या (s>2) / (s<-1) or (s>2)
B (-1<s<2)
C (s=-1) या (s=2) / (s=-1) or (s=2)
D (0<s<1)
Explanation opens after your attempt
Correct Answer
A. (s<-1) या (s>2) / (s<-1) or (s>2)
Step 1
Concept
Here (D=4s-2 -4(s+2)=4(s-2 )(s+1)). From (D>0), (s<-1) or (s>2).
Step 2
Why this answer is correct
The correct answer is A. (s<-1) या (s>2) / (s<-1) or (s>2). Here (D=4s-2 -4(s+2)=4(s-2 )(s+1)). From (D>0), (s<-1) or (s>2).
Step 3
Exam Tip
यहाँ (D=4s-2 -4(s+2)=4(s-2 )(s+1)) है। (D>0) से (s<-1) या (s>2) मिलता है।
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समीकरण \(kx^2-6x+k=0\) के वास्तविक और भिन्न मूलों के लिए सही शर्त क्या है?
What is the correct condition for real and distinct roots of \(kx^2-6x+k=0\)?
#quadratic equations
#real distinct roots
#leading coefficient
A \(k^2<9\) और \(k\neq0\) / \(k^2<9\) and \(k\neq0\)
B \(k^2>9\)
C \(k^2=9\)
D (k=0)
Explanation opens after your attempt
Correct Answer
A. \(k^2<9\) और \(k\neq0\) / \(k^2<9\) and \(k\neq0\)
Step 1
Concept
Here \(D=36-4k^2\). For real and distinct roots (D>0) and \(k\neq0\), hence \(k^2<9\).
Step 2
Why this answer is correct
The correct answer is A. \(k^2<9\) और \(k\neq0\) / \(k^2<9\) and \(k\neq0\). Here \(D=36-4k^2\). For real and distinct roots (D>0) and \(k\neq0\), hence \(k^2<9\).
Step 3
Exam Tip
यहाँ \(D=36-4k^2\) है। वास्तविक और भिन्न मूलों के लिए (D>0) और \(k\neq0\), अतः \(k^2<9\)।
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समीकरण (x-2 -2(k+1)x+k-2 =0) के मूल वास्तविक और भिन्न कब होंगे?
When will the roots of (x-2 -2(k+1)x+k-2 =0) be real and distinct?
#quadratic equations
#real distinct roots
#inequality
A \(k>-\frac{1}{2}\)
B \(k<-\frac{1}{2}\)
C \(k=-\frac{1}{2}\)
D (k=0)
Explanation opens after your attempt
Correct Answer
A. \(k>-\frac{1}{2}\)
Step 1
Concept
Here (D=4(k+1)2 -4k-2 =4(2k+1)). For distinct real roots (D>0), so \(k>-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(k>-\frac{1}{2}\). Here (D=4(k+1)2 -4k-2 =4(2k+1)). For distinct real roots (D>0), so \(k>-\frac{1}{2}\).
Step 3
Exam Tip
यहाँ (D=4(k+1)2 -4k-2 =4(2k+1)) है। भिन्न वास्तविक मूलों के लिए (D>0), इसलिए \(k>-\frac{1}{2}\)।
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यदि \(x^2+px+6=0\) के मूल वास्तविक और भिन्न हैं तो (p) के लिए कौन सी शर्त सही है?
If the roots of \(x^2+px+6=0\) are real and distinct, which condition is correct for (p)?
#quadratic equations
#parameter
#real distinct roots
A \(p^2>24\)
B \(p^2=24\)
C \(p^2<24\)
D \(p^2=6\)
Explanation opens after your attempt
Correct Answer
A. \(p^2>24\)
Step 1
Concept
For real and distinct roots (D>0). So \(p^2-24>0\), that is \(p^2>24\).
Step 2
Why this answer is correct
The correct answer is A. \(p^2>24\). For real and distinct roots (D>0). So \(p^2-24>0\), that is \(p^2>24\).
Step 3
Exam Tip
वास्तविक और भिन्न मूलों के लिए (D>0) होता है। इसलिए \(p^2-24>0\), अर्थात \(p^2>24\)।
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यदि \(x^2+px+4=0\) के मूल वास्तविक और भिन्न हैं तो (p) के लिए सही शर्त क्या है?
If the roots of \(x^2+px+4=0\) are real and distinct, what is the correct condition for (p)?
#quadratic equations
#parameter
#real distinct roots
A \(p^2>16\)
B \(p^2=16\)
C \(p^2<16\)
D \(p^2=4\)
Explanation opens after your attempt
Correct Answer
A. \(p^2>16\)
Step 1
Concept
For real and distinct roots (D>0), so \(p^2-16>0\). Therefore \(p^2>16\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(p^2>16\). For real and distinct roots (D>0), so \(p^2-16>0\). Therefore \(p^2>16\) is correct.
Step 3
Exam Tip
वास्तविक और भिन्न मूलों के लिए (D>0) होता है इसलिए \(p^2-16>0\)। अतः \(p^2>16\) सही है।
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यदि \(x^2-2\theta x+3\theta=0\) के दो वास्तविक और असमान मूल हों, तो \(\theta\) पर कौन सी शर्त सही है?
If \(x^2-2\theta x+3\theta=0\) has two real and distinct roots, which condition on \(\theta\) is correct?
#quadratic-equations
#parameter-inequality
#distinct-roots
A \(\theta<0\) या \(\theta>3\) / \(\theta<0\) or \(\theta>3\)
B \(0<\theta<3\)
C \(\theta=0\) या \(\theta=3\) / \(\theta=0\) or \(\theta=3\)
D हर \(\theta\) / Every \(\theta\)
Explanation opens after your attempt
Correct Answer
A. \(\theta<0\) या \(\theta>3\) / \(\theta<0\) or \(\theta>3\)
Step 1
Concept
Here (D=4\theta-2 -12\theta=4\theta\(\theta-3\)). From (D>0), \(\theta<0\) or \(\theta>3\).
Step 2
Why this answer is correct
The correct answer is A. \(\theta<0\) या \(\theta>3\) / \(\theta<0\) or \(\theta>3\). Here (D=4\theta-2 -12\theta=4\theta\(\theta-3\)). From (D>0), \(\theta<0\) or \(\theta>3\).
Step 3
Exam Tip
यहाँ (D=4\theta-2 -12\theta=4\theta\(\theta-3\)) है। (D>0) से \(\theta<0\) या \(\theta>3\)।
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समीकरण (x-2 -(t+7)x+7t=0) के दो वास्तविक और असमान मूलों के लिए कौन सी शर्त सही है?
Which condition is correct for two real and distinct roots of (x-2 -(t+7)x+7t=0)?
#quadratic-equations
#parameter
#distinct-roots
A \(t\neq7\)
B (t=7)
C (t>7) मात्र / Only (t>7)
D (t<7) मात्र / Only (t<7)
Explanation opens after your attempt
Correct Answer
A. \(t\neq7\)
Step 1
Concept
Here (D=(t+7)2 -28t=(t-7)2 ). For two distinct roots (D>0), so \(t\neq7\).
Step 2
Why this answer is correct
The correct answer is A. \(t\neq7\). Here (D=(t+7)2 -28t=(t-7)2 ). For two distinct roots (D>0), so \(t\neq7\).
Step 3
Exam Tip
यहाँ (D=(t+7)2 -28t=(t-7)2 ) है। दो असमान मूलों के लिए (D>0), इसलिए \(t\neq7\)।
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यदि \(x^2-2\mu x+2\mu=0\) के दो वास्तविक और असमान मूल हों, तो \(\mu\) पर कौन सी शर्त सही है?
If \(x^2-2\mu x+2\mu=0\) has two real and distinct roots, which condition on \(\mu\) is correct?
#quadratic-equations
#parameter-inequality
#distinct-roots
A \(\mu<0\) या \(\mu>2\) / \(\mu<0\) or \(\mu>2\)
B \(0<\mu<2\)
C \(\mu=0\) या \(\mu=2\) / \(\mu=0\) or \(\mu=2\)
D हर \(\mu\) / Every \(\mu\)
Explanation opens after your attempt
Correct Answer
A. \(\mu<0\) या \(\mu>2\) / \(\mu<0\) or \(\mu>2\)
Step 1
Concept
Here (D=4\mu-2 -8\mu=4\mu\(\mu-2\)). From (D>0), \(\mu<0\) or \(\mu>2\).
Step 2
Why this answer is correct
The correct answer is A. \(\mu<0\) या \(\mu>2\) / \(\mu<0\) or \(\mu>2\). Here (D=4\mu-2 -8\mu=4\mu\(\mu-2\)). From (D>0), \(\mu<0\) or \(\mu>2\).
Step 3
Exam Tip
यहाँ (D=4\mu-2 -8\mu=4\mu\(\mu-2\)) है। (D>0) से \(\mu<0\) या \(\mu>2\)।
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समीकरण (x-2 -(r+5)x+5r=0) के दो वास्तविक और असमान मूलों के लिए कौन सी शर्त सही है?
Which condition is correct for two real and distinct roots of (x-2 -(r+5)x+5r=0)?
#quadratic-equations
#parameter
#distinct-roots
A \(r\neq5\)
B (r=5)
C (r>5) मात्र / Only (r>5)
D (r<5) मात्र / Only (r<5)
Explanation opens after your attempt
Correct Answer
A. \(r\neq5\)
Step 1
Concept
Here (D=(r+5)2 -20r=(r-5)2 ). For two distinct roots (D>0), so \(r\neq5\).
Step 2
Why this answer is correct
The correct answer is A. \(r\neq5\). Here (D=(r+5)2 -20r=(r-5)2 ). For two distinct roots (D>0), so \(r\neq5\).
Step 3
Exam Tip
यहाँ (D=(r+5)2 -20r=(r-5)2 ) है। दो असमान मूलों के लिए (D>0), इसलिए \(r\neq5\)।
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समीकरण (3x-2 -2(2k+1)x+(k+1)2 =0) के दो असमान वास्तविक मूलों के लिए कौन सी शर्त सही है?
Which condition is correct for two distinct real roots of (3x-2 -2(2k+1)x+(k+1)2 =0)?
#quadratic-equations
#distinct-roots
#parameter-interval
A (k<-2) या (k>1) / (k<-2) or (k>1)
B (-2<k<1)
C (k=-2) या (k=1) / (k=-2) or (k=1)
D हर (k) / Every (k)
Explanation opens after your attempt
Correct Answer
A. (k<-2) या (k>1) / (k<-2) or (k>1)
Step 1
Concept
Here (D=4(k-1)(k+2)). From (D>0), we get (k<-2) or (k>1).
Step 2
Why this answer is correct
The correct answer is A. (k<-2) या (k>1) / (k<-2) or (k>1). Here (D=4(k-1)(k+2)). From (D>0), we get (k<-2) or (k>1).
Step 3
Exam Tip
यहाँ (D=4(k-1)(k+2)) है। (D>0) से (k<-2) या (k>1) मिलता है।
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यदि \(x^2-2\lambda x+\lambda=0\) के दो वास्तविक और असमान मूल हों, तो \(\lambda\) पर कौन सी शर्त सही है?
If \(x^2-2\lambda x+\lambda=0\) has two real and distinct roots, which condition on \(\lambda\) is correct?
#quadratic-equations
#parameter-inequality
#distinct-roots
A \(\lambda<0\) या \(\lambda>1\) / \(\lambda<0\) or \(\lambda>1\)
B \(0<\lambda<1\)
C \(\lambda=0\) या \(\lambda=1\) / \(\lambda=0\) or \(\lambda=1\)
D हर वास्तविक \(\lambda\) / Every real \(\lambda\)
Explanation opens after your attempt
Correct Answer
A. \(\lambda<0\) या \(\lambda>1\) / \(\lambda<0\) or \(\lambda>1\)
Step 1
Concept
Here (D=4\lambda-2 -4\lambda=4\lambda\(\lambda-1\)). For distinct real roots (D>0), so \(\lambda<0\) or \(\lambda>1\).
Step 2
Why this answer is correct
The correct answer is A. \(\lambda<0\) या \(\lambda>1\) / \(\lambda<0\) or \(\lambda>1\). Here (D=4\lambda-2 -4\lambda=4\lambda\(\lambda-1\)). For distinct real roots (D>0), so \(\lambda<0\) or \(\lambda>1\).
Step 3
Exam Tip
यहाँ (D=4\lambda-2 -4\lambda=4\lambda\(\lambda-1\)) है। असमान वास्तविक मूलों के लिए (D>0), इसलिए \(\lambda<0\) या \(\lambda>1\)।
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यदि (x-2 -2(a+b)x+(a-b)2 =0) के मूल वास्तविक और असमान हों, तो (a) और (b) के लिए सही शर्त क्या है?
If (x-2 -2(a+b)x+(a-b)2 =0) has real and distinct roots, what is the correct condition for (a) and (b)?
#quadratic-equations
#algebraic-parameter
#distinct-roots
A (ab>0)
B (ab=0)
C (ab<0)
D (a=b)
Explanation opens after your attempt
Step 1
Concept
Here (D=4(a+b)2 -4(a-b)2 =16ab). For distinct real roots (D>0), so (ab>0).
Step 2
Why this answer is correct
The correct answer is A. (ab>0). Here (D=4(a+b)2 -4(a-b)2 =16ab). For distinct real roots (D>0), so (ab>0).
Step 3
Exam Tip
यहाँ (D=4(a+b)2 -4(a-b)2 =16ab) है। असमान वास्तविक मूलों के लिए (D>0), इसलिए (ab>0)।
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समीकरण (x-2 -(m+3)x+3m=0) के दो वास्तविक और असमान मूलों के लिए कौन सी शर्त सही है?
Which condition is correct for two real and distinct roots of (x-2 -(m+3)x+3m=0)?
#quadratic-equations
#parameter
#distinct-roots
A \(m\neq3\)
B (m=3)
C (m>3) मात्र / Only (m>3)
D (m<3) मात्र / Only (m<3)
Explanation opens after your attempt
Correct Answer
A. \(m\neq3\)
Step 1
Concept
Here (D=(m+3)2 -12m=(m-3 )2 ). For two distinct roots (D>0), so \(m\neq3\).
Step 2
Why this answer is correct
The correct answer is A. \(m\neq3\). Here (D=(m+3)2 -12m=(m-3 )2 ). For two distinct roots (D>0), so \(m\neq3\).
Step 3
Exam Tip
यहाँ (D=(m+3)2 -12m=(m-3 )2 ) है। दो असमान मूलों के लिए (D>0), इसलिए \(m\neq3\)।
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यदि (3x-2 +(k-2)x+4=0) के दो असमान वास्तविक मूल हों, तो (k) पर कौन सी शर्त सही है?
If (3x-2 +(k-2)x+4=0) has two distinct real roots, which condition on (k) is correct?
#quadratic-equations
#distinct-roots
#parameter-interval
A \(k<2-4\sqrt{3}\) या \(k>2+4\sqrt{3}\) / \(k<2-4\sqrt{3}\) or \(k>2+4\sqrt{3}\)
B \(2-4\sqrt{3}<k<2+4\sqrt{3}\)
C (k=2) मात्र / Only (k=2)
D \(k=4\sqrt{3}\) मात्र / Only \(k=4\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(k<2-4\sqrt{3}\) या \(k>2+4\sqrt{3}\) / \(k<2-4\sqrt{3}\) or \(k>2+4\sqrt{3}\)
Step 1
Concept
Here (D=(k-2)2 -48). For distinct real roots (D>0), so ((k-2)2 >48).
Step 2
Why this answer is correct
The correct answer is A. \(k<2-4\sqrt{3}\) या \(k>2+4\sqrt{3}\) / \(k<2-4\sqrt{3}\) or \(k>2+4\sqrt{3}\). Here (D=(k-2)2 -48). For distinct real roots (D>0), so ((k-2)2 >48).
Step 3
Exam Tip
यहाँ (D=(k-2)2 -48) है। असमान वास्तविक मूलों के लिए (D>0), इसलिए ((k-2)2 >48)।
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समीकरण \(x^2-16x+k=0\) के दो वास्तविक और असमान मूलों के लिए (k) पर कौन सी शर्त सही है?
Which condition on (k) is correct for two real and distinct roots of \(x^2-16x+k=0\)?
#quadratic-equations
#distinct-roots
#parameter
A (k<64)
B (k=64)
C (k>64)
D (k=16)
Explanation opens after your attempt
Step 1
Concept
Here (D=256-4k). For two distinct real roots (D>0), so (k<64).
Step 2
Why this answer is correct
The correct answer is A. (k<64). Here (D=256-4k). For two distinct real roots (D>0), so (k<64).
Step 3
Exam Tip
यहाँ (D=256-4k) है। दो असमान वास्तविक मूलों के लिए (D>0), इसलिए (k<64)।
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यदि (2x-2 +(k+1)x+3=0) के दो असमान वास्तविक मूल हों, तो (k) पर कौन सी शर्त सही है?
If (2x-2 +(k+1)x+3=0) has two distinct real roots, which condition on (k) is correct?
#quadratic-equations
#distinct-roots
#parameter-interval
A \(k<-1-2\sqrt{6}\) या \(k>-1+2\sqrt{6}\) / \(k<-1-2\sqrt{6}\) or \(k>-1+2\sqrt{6}\)
B \(-1-2\sqrt{6}<k<-1+2\sqrt{6}\)
C (k=-1) मात्र / Only (k=-1)
D \(k=2\sqrt{6}\) मात्र / Only \(k=2\sqrt{6}\)
Explanation opens after your attempt
Correct Answer
A. \(k<-1-2\sqrt{6}\) या \(k>-1+2\sqrt{6}\) / \(k<-1-2\sqrt{6}\) or \(k>-1+2\sqrt{6}\)
Step 1
Concept
Here (D=(k+1)2 -24). For distinct real roots (D>0), so ((k+1)2 >24).
Step 2
Why this answer is correct
The correct answer is A. \(k<-1-2\sqrt{6}\) या \(k>-1+2\sqrt{6}\) / \(k<-1-2\sqrt{6}\) or \(k>-1+2\sqrt{6}\). Here (D=(k+1)2 -24). For distinct real roots (D>0), so ((k+1)2 >24).
Step 3
Exam Tip
यहाँ (D=(k+1)2 -24) है। असमान वास्तविक मूलों के लिए (D>0), इसलिए ((k+1)2 >24)।
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समीकरण \(x^2-12x+k=0\) के दो वास्तविक और असमान मूलों के लिए कौन सी शर्त सही है?
Which condition is correct for two real and distinct roots of \(x^2-12x+k=0\)?
#quadratic-equations
#distinct-roots
#parameter
A (k<36)
B (k=36)
C (k>36)
D (k=12)
Explanation opens after your attempt
Step 1
Concept
Here (D=144-4k). For two distinct real roots (D>0), so (k<36).
Step 2
Why this answer is correct
The correct answer is A. (k<36). Here (D=144-4k). For two distinct real roots (D>0), so (k<36).
Step 3
Exam Tip
यहाँ (D=144-4k) है। दो असमान वास्तविक मूलों के लिए (D>0), इसलिए (k<36)।
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कौन सा समीकरण वास्तविक, अपरिमेय और भिन्न मूल देता है?
Which equation gives real, irrational and distinct roots?
#quadratic equations
#irrational distinct roots
#choose equation
A \(x^2-10x+23=0\)
B \(x^2-10x+24=0\)
C \(x^2-10x+25=0\)
D \(x^2+10x+26=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-10x+23=0\)
Step 1
Concept
In the first equation, (D=100-92=8>0), and (8) is not a perfect square. So the roots are real, irrational and distinct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+23=0\). In the first equation, (D=100-92=8>0), and (8) is not a perfect square. So the roots are real, irrational and distinct.
Step 3
Exam Tip
पहले समीकरण में (D=100-92=8>0) है और (8) पूर्ण वर्ग नहीं है। इसलिए मूल वास्तविक, अपरिमेय और भिन्न हैं।
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समीकरण (x-2 +2(k+1)x+k-2 =0) के दो असमान वास्तविक मूलों के लिए सही शर्त चुनिए।
Choose the correct condition for two distinct real roots of (x-2 +2(k+1)x+k-2 =0).
#quadratic-equations
#distinct-roots
#parameter
A \(k>-\frac{1}{2}\)
B \(k=-\frac{1}{2}\)
C \(k<-\frac{1}{2}\)
D (k=0) मात्र / Only (k=0)
Explanation opens after your attempt
Correct Answer
A. \(k>-\frac{1}{2}\)
Step 1
Concept
Here (D=4(k+1)2 -4k-2 =4(2k+1)). For distinct real roots (D>0), so \(k>-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(k>-\frac{1}{2}\). Here (D=4(k+1)2 -4k-2 =4(2k+1)). For distinct real roots (D>0), so \(k>-\frac{1}{2}\).
Step 3
Exam Tip
यहाँ (D=4(k+1)2 -4k-2 =4(2k+1)) है। असमान वास्तविक मूलों के लिए (D>0), इसलिए \(k>-\frac{1}{2}\)।
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यदि किसी द्विघात समीकरण के दो असमान वास्तविक मूल हैं, तो (D) कैसा होगा?
If a quadratic equation has two distinct real roots, how will (D) be?
#quadratic equations
#concept
#distinct roots
A (D>0)
B (D=0)
C (D<0)
D \(D\leq0\)
Explanation opens after your attempt
Step 1
Concept
For distinct real roots, (D>0). Do not add the equality sign by mistake.
Step 2
Why this answer is correct
The correct answer is A. (D>0). For distinct real roots, (D>0). Do not add the equality sign by mistake.
Step 3
Exam Tip
असमान वास्तविक मूलों के लिए (D>0) होता है। बराबर का चिन्ह गलती से न लगाएं।
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समीकरण \(2x^2+3x+\lambda=0\) के वास्तविक और असमान मूलों के लिए कौन सी शर्त सही है?
For \(2x^2+3x+\lambda=0\) to have real and distinct roots, which condition is correct?
#quadratic equations
#parameter
#distinct roots
#fraction
A \(\lambda<\frac{9}{8}\)
B \(\lambda=\frac{9}{8}\)
C \(\lambda>\frac{9}{8}\)
D \(\lambda=\frac{8}{9}\)
Explanation opens after your attempt
Correct Answer
A. \(\lambda<\frac{9}{8}\)
Step 1
Concept
(D=32 -4(2)\lambda=9-8\lambda). From (D>0), we get \(\lambda<\frac{9}{8}\).
Step 2
Why this answer is correct
The correct answer is A. \(\lambda<\frac{9}{8}\). (D=32 -4(2)\lambda=9-8\lambda). From (D>0), we get \(\lambda<\frac{9}{8}\).
Step 3
Exam Tip
(D=32 -4(2)\lambda=9-8\lambda) है। (D>0) से \(\lambda<\frac{9}{8}\) मिलता है।
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समीकरण \(x^2-2x+n=0\) के दो वास्तविक और असमान मूल होने के लिए कौन सी शर्त सही है?
For \(x^2-2x+n=0\) to have two real and distinct roots, which condition is correct?
#quadratic equations
#parameter
#distinct roots
#inequality
A (n<1)
B (n=1)
C (n>1)
D (n=2)
Explanation opens after your attempt
Step 1
Concept
For distinct real roots (D>0), so ((-2)2 -4n>0) gives (n<1). Use a strict inequality for distinct roots.
Step 2
Why this answer is correct
The correct answer is A. (n<1). For distinct real roots (D>0), so ((-2)2 -4n>0) gives (n<1). Use a strict inequality for distinct roots.
Step 3
Exam Tip
असमान वास्तविक मूलों के लिए (D>0), इसलिए ((-2)2 -4n>0) से (n<1)। असमान के लिए कड़ाई वाली असमता लगती है।
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यदि (D=0) और \(a\neq0\) हो तो द्विघात समीकरण में कितने अलग-अलग वास्तविक मूल होंगे?
If (D=0) and \(a\neq0\), how many distinct real roots will the quadratic equation have?
#quadratic equations
#equal roots
#distinct count
A (1)
B (2)
C (0)
D (3)
Explanation opens after your attempt
Step 1
Concept
At (D=0), both roots are equal, so the number of distinct real roots is (1). Remember the root is repeated.
Step 2
Why this answer is correct
The correct answer is A. (1). At (D=0), both roots are equal, so the number of distinct real roots is (1). Remember the root is repeated.
Step 3
Exam Tip
(D=0) पर दोनों मूल समान होते हैं, इसलिए अलग-अलग वास्तविक मूलों की संख्या (1) है। ध्यान रखें मूल दो बार दोहरता है।
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किस समीकरण के दो वास्तविक और असमान मूल होंगे?
Which equation will have two real and distinct roots?
#quadratic equations
#choose equation
#distinct roots
A \(x^2-7x+10=0\)
B \(x^2+2x+1=0\)
C \(x^2+4x+8=0\)
D \(4x^2+4x+1=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+10=0\)
Step 1
Concept
For the first equation, (D=(-7)2 -4(1)(10)=9>0). Hence its roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-7x+10=0\). For the first equation, (D=(-7)2 -4(1)(10)=9>0). Hence its roots are real and distinct.
Step 3
Exam Tip
पहले समीकरण में (D=(-7)2 -4(1)(10)=9>0) है। इसलिए उसके मूल वास्तविक और असमान हैं।
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\(x^2-10x+k=0\) की जड़ें भिन्न अभाज्य संख्याएँ हैं, तो (k) का मान क्या है?
The roots of \(x^2-10x+k=0\) are distinct prime numbers. What is (k)?
#quadratic-roots
#prime-roots
#integer-roots
A (21)
B (25)
C (16)
D (10)
Explanation opens after your attempt
Step 1
Concept
The distinct prime roots with sum (10) are (3) and (7). Their product is (21), so (k=21).
Step 2
Why this answer is correct
The correct answer is A. (21). The distinct prime roots with sum (10) are (3) and (7). Their product is (21), so (k=21).
Step 3
Exam Tip
योग (10) वाली भिन्न अभाज्य जड़ें (3) और (7) हैं। उनका गुणनफल (21) है, इसलिए (k=21)।
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समीकरण (3x-2 -2(2a+1)x+\(a^2+a+1\)=0) के वास्तविक और भिन्न मूल कब होंगे?
When will (3x-2 -2(2a+1)x+\(a^2+a+1\)=0) have real and distinct roots?
#quadratic equations
#real distinct
#interval
A (a<-2) या (a>1) / (a<-2) or (a>1)
B (-2<a<1)
C (a=-2) या (a=1) / (a=-2) or (a=1)
D सभी वास्तविक (a) / All real (a)
Explanation opens after your attempt
Correct Answer
A. (a<-2) या (a>1) / (a<-2) or (a>1)
Step 1
Concept
For real and distinct roots, (D>0) is needed. From \(a^2+a-2>0\), we get (a<-2) or (a>1).
Step 2
Why this answer is correct
The correct answer is A. (a<-2) या (a>1) / (a<-2) or (a>1). For real and distinct roots, (D>0) is needed. From \(a^2+a-2>0\), we get (a<-2) or (a>1).
Step 3
Exam Tip
वास्तविक और भिन्न मूलों के लिए (D>0) चाहिए। \(a^2+a-2>0\) से (a<-2) या (a>1) मिलता है।
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यदि \(x^2-2hx+h^2+8h=0\) के मूल वास्तविक और भिन्न हैं, तो (h) पर सही शर्त क्या है?
If \(x^2-2hx+h^2+8h=0\) has real and distinct roots, what is the correct condition on (h)?
#quadratic equations
#parameter inequality
#real distinct
A (h<0)
B (h>0)
C (h=0)
D \(h\ge0\)
Explanation opens after your attempt
Step 1
Concept
Here (D=4h-2 -4\(h^2+8h\)=-32h). For (D>0), (h<0) is required.
Step 2
Why this answer is correct
The correct answer is A. (h<0). Here (D=4h-2 -4\(h^2+8h\)=-32h). For (D>0), (h<0) is required.
Step 3
Exam Tip
यहाँ (D=4h-2 -4\(h^2+8h\)=-32h) है। (D>0) के लिए (h<0) चाहिए।
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यदि किसी द्विघात का विविक्तकर (D=20n-80) है, तो दो वास्तविक और असमान मूलों के लिए (n) पर कौन सी शर्त होगी?
If a quadratic has discriminant (D=20n-80), what condition on (n) gives two real and distinct roots?
#quadratic-equations
#discriminant-inequality
#distinct-roots
A (n>4)
B (n=4)
C (n<4)
D हर (n) / Every (n)
Explanation opens after your attempt
Step 1
Concept
For two distinct real roots (D>0) is needed. (20n-80>0) gives (n>4).
Step 2
Why this answer is correct
The correct answer is A. (n>4). For two distinct real roots (D>0) is needed. (20n-80>0) gives (n>4).
Step 3
Exam Tip
दो असमान वास्तविक मूलों के लिए (D>0) चाहिए। (20n-80>0) से (n>4)।
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यदि किसी द्विघात का विविक्तकर (D=12n-36) है, तो दो वास्तविक और असमान मूलों के लिए (n) पर कौन सी शर्त होगी?
If a quadratic has discriminant (D=12n-36), what condition on (n) gives two real and distinct roots?
#quadratic-equations
#discriminant-inequality
#distinct-roots
A (n>3)
B (n=3)
C (n<3)
D हर (n) / Every (n)
Explanation opens after your attempt
Step 1
Concept
For two distinct real roots (D>0) is needed. (12n-36>0) gives (n>3).
Step 2
Why this answer is correct
The correct answer is A. (n>3). For two distinct real roots (D>0) is needed. (12n-36>0) gives (n>3).
Step 3
Exam Tip
दो असमान वास्तविक मूलों के लिए (D>0) चाहिए। (12n-36>0) से (n>3) मिलता है।
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समीकरण (2x-2 -(3q+1)x+q=0) के मूल हमेशा वास्तविक और भिन्न क्यों हैं?
Why are the roots of (2x-2 -(3q+1)x+q=0) always real and distinct?
#quadratic equations
#always real roots
#reasoning
A क्योंकि \(D=9q^2-2q+1>0\) / Because \(D=9q^2-2q+1>0\)
B क्योंकि (D=0) / Because (D=0)
C क्योंकि (D<0) / Because (D<0)
D क्योंकि (a=0) / Because (a=0)
Explanation opens after your attempt
Correct Answer
A. क्योंकि \(D=9q^2-2q+1>0\) / Because \(D=9q^2-2q+1>0\)
Step 1
Concept
Here (D=(3q+1)2 -8q=9q-2 -2q+1). Its own discriminant ((-2)2 -4(9)(1)<0), so it is always positive.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(D=9q^2-2q+1>0\) / Because \(D=9q^2-2q+1>0\). Here (D=(3q+1)2 -8q=9q-2 -2q+1). Its own discriminant ((-2)2 -4(9)(1)<0), so it is always positive.
Step 3
Exam Tip
यहाँ (D=(3q+1)2 -8q=9q-2 -2q+1) है। इसका अपना विविक्तकर ((-2)2 -4(9)(1)<0) और मान सदैव धनात्मक है।
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कथन: \(x^2+3x+7=0\) के वास्तविक मूल नहीं हैं। कारण: (D<0) होने पर वास्तविक मूल नहीं होते। सही विकल्प चुनिए।
Assertion: \(x^2+3x+7=0\) has no real roots. Reason: When (D<0), real roots do not exist. Choose the correct option.
#quadratic-equations
#assertion-reason
#no-real-roots
A कथन और कारण दोनों सही हैं / Both assertion and reason are correct
B कथन सही है, कारण गलत है / Assertion is correct, reason is wrong
C कथन गलत है, कारण सही है / Assertion is wrong, reason is correct
D कथन और कारण दोनों गलत हैं / Both assertion and reason are wrong
Explanation opens after your attempt
Correct Answer
A. कथन और कारण दोनों सही हैं / Both assertion and reason are correct
Step 1
Concept
Here (D=32 -4(1)(7)=-19). Since (D<0), the assertion is correct.
Step 2
Why this answer is correct
The correct answer is A. कथन और कारण दोनों सही हैं / Both assertion and reason are correct. Here (D=32 -4(1)(7)=-19). Since (D<0), the assertion is correct.
Step 3
Exam Tip
यहाँ (D=32 -4(1)(7)=-19) है। (D<0) होने से कथन सही है।
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निम्न में से किस समीकरण के मूल वास्तविक, अपरिमेय और असमान हैं?
Which of the following equations has real, irrational, and distinct roots?
#quadratic-equations
#choose-equation
#irrational-roots
A \(x^2-2\sqrt{2}x-1=0\)
B \(x^2-4x+4=0\)
C \(x^2+2x+5=0\)
D \(x^2-5x+6=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-2\sqrt{2}x-1=0\)
Step 1
Concept
In option (A), (D=\(-2\sqrt{2}\)2 -4(1)(-1)=12). (12) is positive but not a perfect square.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-2\sqrt{2}x-1=0\). In option (A), (D=\(-2\sqrt{2}\)2 -4(1)(-1)=12). (12) is positive but not a perfect square.
Step 3
Exam Tip
विकल्प (A) में (D=\(-2\sqrt{2}\)2 -4(1)(-1)=12) है। (12) धनात्मक है पर पूर्ण वर्ग नहीं है।
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निम्न में से किस समीकरण के दो वास्तविक परिमेय और असमान मूल हैं?
Which of the following equations has two real rational and distinct roots?
#quadratic-equations
#choose-equation
#rational-roots
A \(x^2-17x+72=0\)
B \(x^2-17x+80=0\)
C \(x^2+17x+80=0\) जहाँ (D=-31) / \(x^2+17x+80=0\) with (D=-31)
D \(x^2-18x+81=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-17x+72=0\)
Step 1
Concept
In option (A), (D=(-17)2 -4(1)(72)=1). A positive perfect-square (D) gives rational distinct roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-17x+72=0\). In option (A), (D=(-17)2 -4(1)(72)=1). A positive perfect-square (D) gives rational distinct roots.
Step 3
Exam Tip
विकल्प (A) में (D=(-17)2 -4(1)(72)=1) है। धनात्मक पूर्ण वर्ग (D) परिमेय असमान मूल देता है।
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कौन सा समीकरण वास्तविक, अपरिमेय और भिन्न मूल रखता है?
Which equation has real, irrational and distinct roots?
#quadratic equations
#choose equation
#irrational roots
A \(x^2-2x-3=0\)
B \(x^2-2x-2=0\)
C \(x^2-2x+1=0\)
D \(x^2+2x+5=0\)
Explanation opens after your attempt
Correct Answer
B. \(x^2-2x-2=0\)
Step 1
Concept
In the second equation (D=(-2)2 -4(1)(-2)=12). (12) is positive but not a perfect square.
Step 2
Why this answer is correct
The correct answer is B. \(x^2-2x-2=0\). In the second equation (D=(-2)2 -4(1)(-2)=12). (12) is positive but not a perfect square.
Step 3
Exam Tip
दूसरे समीकरण में (D=(-2)2 -4(1)(-2)=12) है। (12) धनात्मक है पर पूर्ण वर्ग नहीं है।
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कौन सा समीकरण वास्तविक, परिमेय और भिन्न मूल रखता है?
Which equation has real, rational and distinct roots?
#quadratic equations
#choose equation
#rational roots
A \(x^2-9x+20=0\)
B \(x^2-9x+21=0\)
C \(x^2+9x+30=0\)
D \(x^2+2x+5=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-9x+20=0\)
Step 1
Concept
In the first equation (D=(-9)2 -4(1)(20)=1). Hence the roots are real, rational and distinct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-9x+20=0\). In the first equation (D=(-9)2 -4(1)(20)=1). Hence the roots are real, rational and distinct.
Step 3
Exam Tip
पहले समीकरण में (D=(-9)2 -4(1)(20)=1) है। इसलिए मूल वास्तविक, परिमेय और भिन्न हैं।
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किस स्थिति में द्विघात समीकरण के दो भिन्न वास्तविक मूल होते हैं?
In which condition does a quadratic equation have two distinct real roots?
#roots
#discriminant
#distinct_roots
A (D=0)
B (D<0)
C (D>0)
D (a=0)
Explanation opens after your attempt
Step 1
Concept
For two distinct real roots, (D>0) is required. This is a direct rule to check the nature.
Step 2
Why this answer is correct
The correct answer is C. (D>0). For two distinct real roots, (D>0) is required. This is a direct rule to check the nature.
Step 3
Exam Tip
दो भिन्न वास्तविक मूलों के लिए (D>0) होना चाहिए। यह प्रकृति जांचने का सीधा नियम है।
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समीकरण \(x^2-20x+k=0\) के मूल वास्तविक और भिन्न हों, तो (k) पर सही शर्त क्या है?
If the roots of \(x^2-20x+k=0\) are real and distinct, what is the correct condition on (k)?
#quadratic-equations
#distinct-real-roots
#parameter
#expert
A (k<100)
B (k=100)
C (k>100)
D \(k\leq100\)
Explanation opens after your attempt
Correct Answer
A. (k<100)
Step 1
Concept
For real and distinct roots, (D>0) is needed. Here (400-4k>0), so (k<100).
Step 2
Why this answer is correct
The correct answer is A. (k<100). For real and distinct roots, (D>0) is needed. Here (400-4k>0), so (k<100).
Step 3
Exam Tip
भिन्न वास्तविक मूलों के लिए (D>0) चाहिए। यहाँ (400-4k>0), इसलिए (k<100)।
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समीकरण \(x^2-16x+k=0\) के मूल वास्तविक और भिन्न हों, तो (k) पर सही शर्त क्या है?
If the roots of \(x^2-16x+k=0\) are real and distinct, what is the correct condition on (k)?
#quadratic-equations
#distinct-real-roots
#parameter
#expert
A (k<64)
B (k=64)
C (k>64)
D \(k\leq64\)
Explanation opens after your attempt
Step 1
Concept
For real and distinct roots, (D>0) is needed. Here (256-4k>0), so (k<64).
Step 2
Why this answer is correct
The correct answer is A. (k<64). For real and distinct roots, (D>0) is needed. Here (256-4k>0), so (k<64).
Step 3
Exam Tip
भिन्न वास्तविक मूलों के लिए (D>0) चाहिए। यहाँ (256-4k>0), इसलिए (k<64)।
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समीकरण \(x^2-12x+k=0\) के मूल वास्तविक और भिन्न हों, तो (k) पर सही शर्त क्या है?
If the roots of \(x^2-12x+k=0\) are real and distinct, what is the correct condition on (k)?
#quadratic-equations
#distinct-real-roots
#parameter
#expert
A (k<36)
B (k=36)
C (k>36)
D \(k\leq36\)
Explanation opens after your attempt
Step 1
Concept
For real and distinct roots, (D>0) is required. Here (144-4k>0), so (k<36).
Step 2
Why this answer is correct
The correct answer is A. (k<36). For real and distinct roots, (D>0) is required. Here (144-4k>0), so (k<36).
Step 3
Exam Tip
भिन्न वास्तविक मूलों के लिए (D>0) चाहिए। यहाँ (144-4k>0), इसलिए (k<36)।
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समीकरण \(x^2-8x+k=0\) के मूल वास्तविक और भिन्न हों, तो (k) पर सही शर्त क्या है?
If the roots of \(x^2-8x+k=0\) are real and distinct, what is the correct condition on (k)?
#quadratic-equations
#distinct-real-roots
#parameter
#hard
A (k<16)
B (k=16)
C (k>16)
D \(k\leq16\)
Explanation opens after your attempt
Step 1
Concept
For real and distinct roots, (D>0) is needed. Here (64-4k>0), so (k<16).
Step 2
Why this answer is correct
The correct answer is A. (k<16). For real and distinct roots, (D>0) is needed. Here (64-4k>0), so (k<16).
Step 3
Exam Tip
भिन्न वास्तविक मूलों के लिए (D>0) चाहिए। यहाँ (64-4k>0), इसलिए (k<16)।
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समीकरण \(x^2-6x+k=0\) के मूल वास्तविक और भिन्न हों, तो (k) पर सही शर्त क्या है?
If the roots of \(x^2-6x+k=0\) are real and distinct, what is the correct condition on (k)?
#quadratic-equations
#distinct-real-roots
#parameter
#hard
A (k<9)
B (k=9)
C (k>9)
D \(k\leq9\)
Explanation opens after your attempt
Step 1
Concept
For real and distinct roots, (D>0) is needed. Here (36-4k>0), so (k<9).
Step 2
Why this answer is correct
The correct answer is A. (k<9). For real and distinct roots, (D>0) is needed. Here (36-4k>0), so (k<9).
Step 3
Exam Tip
भिन्न वास्तविक मूलों के लिए (D>0) चाहिए। यहाँ (36-4k>0), इसलिए (k<9)।
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यदि \(kx^2-12x+k=0\) की जड़ें वास्तविक और व्युत्क्रम हैं, तो (k) पर सही शर्त क्या है?
If \(kx^2-12x+k=0\) has real reciprocal roots, what is the correct condition on (k)?
#quadratic-roots
#reciprocal-roots
#real-roots
A \(k\neq0\) और \(k^2\le36\) / \(k\neq0\) and \(k^2\le36\)
B (k=0)
C \(k^2>36\)
D (k=12) केवल / (k=12) only
Explanation opens after your attempt
Correct Answer
A. \(k\neq0\) और \(k^2\le36\) / \(k\neq0\) and \(k^2\le36\)
Step 1
Concept
The product of roots is \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(144-4k^2\ge0\), hence \(k^2\le36\).
Step 2
Why this answer is correct
The correct answer is A. \(k\neq0\) और \(k^2\le36\) / \(k\neq0\) and \(k^2\le36\). The product of roots is \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(144-4k^2\ge0\), hence \(k^2\le36\).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{k}{k}=1\) है, इसलिए \(k\neq0\) चाहिए। वास्तविक जड़ों के लिए \(144-4k^2\ge0\), अतः \(k^2\le36\)।
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यदि \(kx^2-10x+k=0\) की जड़ें वास्तविक और व्युत्क्रम हैं, तो (k) पर सही शर्त कौन-सी है?
If \(kx^2-10x+k=0\) has real reciprocal roots, which condition on (k) is correct?
#quadratic-roots
#reciprocal-roots
#real-roots
A (k=0)
B \(k^2>25\)
C \(k\neq0\) और \(k^2\le25\) / \(k\neq0\) and \(k^2\le25\)
D (k=10) केवल / (k=10) only
Explanation opens after your attempt
Correct Answer
C. \(k\neq0\) और \(k^2\le25\) / \(k\neq0\) and \(k^2\le25\)
Step 1
Concept
The product of roots is \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(100-4k^2\ge0\), hence \(k^2\le25\).
Step 2
Why this answer is correct
The correct answer is C. \(k\neq0\) और \(k^2\le25\) / \(k\neq0\) and \(k^2\le25\). The product of roots is \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(100-4k^2\ge0\), hence \(k^2\le25\).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{k}{k}=1\) है, इसलिए \(k\neq0\) चाहिए। वास्तविक जड़ों के लिए \(100-4k^2\ge0\), अतः \(k^2\le25\)।
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यदि \(kx^2-8x+k=0\) की जड़ें वास्तविक और व्युत्क्रम हैं, तो (k) पर सही शर्त क्या है?
If the roots of \(kx^2-8x+k=0\) are real and reciprocal, what is the correct condition on (k)?
#quadratic-roots
#reciprocal-roots
#real-roots
A \(k\neq0\) और \(k^2\le16\) / \(k\neq0\) and \(k^2\le16\)
B (k=0)
C \(k^2>16\)
D (k=8) केवल / (k=8) only
Explanation opens after your attempt
Correct Answer
A. \(k\neq0\) और \(k^2\le16\) / \(k\neq0\) and \(k^2\le16\)
Step 1
Concept
For reciprocal roots, \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(64-4k^2\ge0\), hence \(k^2\le16\).
Step 2
Why this answer is correct
The correct answer is A. \(k\neq0\) और \(k^2\le16\) / \(k\neq0\) and \(k^2\le16\). For reciprocal roots, \(\frac{k}{k}=1\), so \(k\neq0\) is needed. For real roots, \(64-4k^2\ge0\), hence \(k^2\le16\).
Step 3
Exam Tip
व्युत्क्रम जड़ों के लिए \(\frac{k}{k}=1\) है, इसलिए \(k\neq0\) चाहिए। वास्तविक जड़ों के लिए \(64-4k^2\ge0\), अतः \(k^2\le16\)।
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\(x^2-4x+k=0\) की जड़ें वास्तविक और व्युत्क्रम हों, तो (k) का मान क्या है?
If the roots of \(x^2-4x+k=0\) are real and reciprocal, what is (k)?
#quadratic-roots
#reciprocal-roots
#real-roots
A (1)
B (2)
C (4)
D (-1)
Explanation opens after your attempt
Step 1
Concept
For reciprocal roots, \(\alpha\beta=1\). Here \(\alpha\beta=k\), so (k=1), and (D=12>0) confirms real roots.
Step 2
Why this answer is correct
The correct answer is A. (1). For reciprocal roots, \(\alpha\beta=1\). Here \(\alpha\beta=k\), so (k=1), and (D=12>0) confirms real roots.
Step 3
Exam Tip
व्युत्क्रम जड़ों के लिए \(\alpha\beta=1\) होता है। यहाँ \(\alpha\beta=k\), इसलिए (k=1), और (D=12>0) से जड़ें वास्तविक भी हैं।
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एक संख्या समस्या से समीकरण (n-2 -2pn+\(p^2-11p\)=0) बनता है। दो वास्तविक और असमान (n) के लिए (p) पर कौन सी शर्त है?
A number problem gives (n-2 -2pn+\(p^2-11p\)=0). What condition on (p) gives two real and distinct values of (n)?
#quadratic-equations
#application
#distinct-roots
A (p>0)
B (p=0)
C (p<0)
D हर (p) / Every (p)
Explanation opens after your attempt
Step 1
Concept
Here (D=4p-2 -4\(p^2-11p\)=44p). For two distinct real values (D>0), so (p>0).
Step 2
Why this answer is correct
The correct answer is A. (p>0). Here (D=4p-2 -4\(p^2-11p\)=44p). For two distinct real values (D>0), so (p>0).
Step 3
Exam Tip
यहाँ (D=4p-2 -4\(p^2-11p\)=44p) है। दो असमान वास्तविक मानों के लिए (D>0), इसलिए (p>0)।
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एक संख्या समस्या से समीकरण (n-2 -2pn+\(p^2-7p\)=0) बनता है। दो वास्तविक और असमान (n) के लिए (p) पर कौन सी शर्त है?
A number problem gives (n-2 -2pn+\(p^2-7p\)=0). What condition on (p) gives two real and distinct values of (n)?
#quadratic-equations
#application
#distinct-roots
A (p>0)
B (p=0)
C (p<0)
D हर (p) / Every (p)
Explanation opens after your attempt
Step 1
Concept
Here (D=4p-2 -4\(p^2-7p\)=28p). For two distinct real values (D>0), so (p>0).
Step 2
Why this answer is correct
The correct answer is A. (p>0). Here (D=4p-2 -4\(p^2-7p\)=28p). For two distinct real values (D>0), so (p>0).
Step 3
Exam Tip
यहाँ (D=4p-2 -4\(p^2-7p\)=28p) है। दो असमान वास्तविक मानों के लिए (D>0), इसलिए (p>0)।
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एक संख्या पहेली से समीकरण (n-2 -2pn+\(p^2-5p\)=0) बनता है। दो वास्तविक और असमान (n) के लिए (p) पर कौन सी शर्त है?
A number puzzle gives (n-2 -2pn+\(p^2-5p\)=0). What condition on (p) gives two real and distinct values of (n)?
#quadratic-equations
#application
#distinct-roots
A (p>0)
B (p=0)
C (p<0)
D हर (p) / Every (p)
Explanation opens after your attempt
Step 1
Concept
Here (D=4p-2 -4\(p^2-5p\)=20p). For two distinct real values (D>0), so (p>0).
Step 2
Why this answer is correct
The correct answer is A. (p>0). Here (D=4p-2 -4\(p^2-5p\)=20p). For two distinct real values (D>0), so (p>0).
Step 3
Exam Tip
यहाँ (D=4p-2 -4\(p^2-5p\)=20p) है। दो असमान वास्तविक मानों के लिए (D>0), इसलिए (p>0)।
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यदि \(x^2-16x+n=0\) के दो अलग वास्तविक मूल हैं, तो (n) के लिए कौनसी शर्त सही है?
If \(x^2-16x+n=0\) has two distinct real roots, which condition on (n) is correct?
#quadratic
#discriminant
#distinct-roots
A (n<64)
B (n>64)
C (n=64)
D \(n\ge64\)
Explanation opens after your attempt
Step 1
Concept
For two distinct real roots, (D>0), so (256-4n>0) and (n<64). In exams, connect (D>0) with distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. (n<64). For two distinct real roots, (D>0), so (256-4n>0) and (n<64). In exams, connect (D>0) with distinct real roots.
Step 3
Exam Tip
दो अलग वास्तविक मूलों के लिए (D>0), इसलिए (256-4n>0) और (n<64) है। परीक्षा में (D>0) को अलग वास्तविक मूल से जोड़ें।
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यदि \(x^2-14x+n=0\) के दो अलग वास्तविक मूल हैं, तो (n) के लिए कौनसी शर्त सही है?
If \(x^2-14x+n=0\) has two distinct real roots, which condition on (n) is correct?
#quadratic
#discriminant
#distinct-roots
A (n<49)
B (n>49)
C (n=49)
D \(n\ge49\)
Explanation opens after your attempt
Step 1
Concept
For two distinct real roots, (D>0), so (196-4n>0) and (n<49). In exams, connect (D>0) with distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. (n<49). For two distinct real roots, (D>0), so (196-4n>0) and (n<49). In exams, connect (D>0) with distinct real roots.
Step 3
Exam Tip
दो अलग वास्तविक मूलों के लिए (D>0), इसलिए (196-4n>0) और (n<49) है। परीक्षा में (D>0) को अलग वास्तविक मूल से जोड़ें।
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यदि \(x^2-12x+n=0\) के दो अलग वास्तविक मूल हैं, तो (n) के लिए कौनसी शर्त सही है?
If \(x^2-12x+n=0\) has two distinct real roots, which condition on (n) is correct?
#quadratic
#discriminant
#distinct-roots
A (n<36)
B (n>36)
C (n=36)
D \(n\ge36\)
Explanation opens after your attempt
Step 1
Concept
For two distinct real roots, (D>0), so (144-4n>0) and (n<36). In exams, connect (D>0) with distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. (n<36). For two distinct real roots, (D>0), so (144-4n>0) and (n<36). In exams, connect (D>0) with distinct real roots.
Step 3
Exam Tip
दो अलग वास्तविक मूलों के लिए (D>0), इसलिए (144-4n>0) और (n<36) है। परीक्षा में (D>0) को अलग वास्तविक मूल से जोड़ें।
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यदि \(x^2-10x+n=0\) के दो अलग वास्तविक मूल हैं, तो (n) के लिए कौनसी शर्त सही है?
If \(x^2-10x+n=0\) has two distinct real roots, which condition on (n) is correct?
#quadratic
#discriminant
#distinct-roots
A (n<25)
B (n>25)
C (n=25)
D \(n\ge25\)
Explanation opens after your attempt
Step 1
Concept
For two distinct real roots, (D>0), so (100-4n>0) and (n<25). In exams, connect (D>0) with distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. (n<25). For two distinct real roots, (D>0), so (100-4n>0) and (n<25). In exams, connect (D>0) with distinct real roots.
Step 3
Exam Tip
दो अलग वास्तविक मूलों के लिए (D>0), इसलिए (100-4n>0) और (n<25) है। परीक्षा में (D>0) को अलग वास्तविक मूल से जोड़ें।
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यदि \(x^2-8x+n=0\) के दो अलग वास्तविक मूल हैं, तो (n) के लिए कौनसी शर्त सही है?
If \(x^2-8x+n=0\) has two distinct real roots, which condition on (n) is correct?
#quadratic
#discriminant
#distinct-roots
A (n<16)
B (n>16)
C (n=16)
D \(n\ge16\)
Explanation opens after your attempt
Step 1
Concept
For two distinct real roots, (D>0), so (64-4n>0) and (n<16). In exams, connect (D>0) with distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. (n<16). For two distinct real roots, (D>0), so (64-4n>0) and (n<16). In exams, connect (D>0) with distinct real roots.
Step 3
Exam Tip
दो अलग वास्तविक मूलों के लिए (D>0), इसलिए (64-4n>0) और (n<16) है। परीक्षा में (D>0) को अलग वास्तविक मूल से जोड़ें।
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यदि \(x^2-4x+n=0\) के दो अलग वास्तविक मूल हैं, तो (n) के लिए कौनसी शर्त सही है?
If \(x^2-4x+n=0\) has two distinct real roots, which condition on (n) is correct?
#quadratic
#discriminant
#distinct-roots
A (n<4)
B (n>4)
C (n=4)
D \(n\ge4\)
Explanation opens after your attempt
Step 1
Concept
For two distinct real roots, (D>0), so (16-4n>0) and (n<4). In exams, connect (D>0) with distinct roots.
Step 2
Why this answer is correct
The correct answer is A. (n<4). For two distinct real roots, (D>0), so (16-4n>0) and (n<4). In exams, connect (D>0) with distinct roots.
Step 3
Exam Tip
दो अलग वास्तविक मूलों के लिए (D>0), इसलिए (16-4n>0) और (n<4) है। परीक्षा में (D>0) को distinct roots से जोड़ें।
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यदि \(x^2-6x+c=0\) की जड़ें \(\alpha,\beta\) हैं और \(\alpha^2+\beta^2=26\), तो जड़ें क्या हैं?
If \(\alpha,\beta\) are roots of \(x^2-6x+c=0\) and \(\alpha^2+\beta^2=26\), what are the roots?
#quadratic-roots
#determine-roots
#sum-product
A (1) और (5) / (1) and (5)
B (2) और (4) / (2) and (4)
C (3) और (3) / (3) and (3)
D (0) और (6) / (0) and (6)
Explanation opens after your attempt
Correct Answer
A. (1) और (5) / (1) and (5)
Step 1
Concept
Here \(\alpha+\beta=6\) and \(\alpha^2+\beta^2=26\). From \(36-2\alpha\beta=26\), \(\alpha\beta=5\), so the roots are (1) and (5).
Step 2
Why this answer is correct
The correct answer is A. (1) और (5) / (1) and (5). Here \(\alpha+\beta=6\) and \(\alpha^2+\beta^2=26\). From \(36-2\alpha\beta=26\), \(\alpha\beta=5\), so the roots are (1) and (5).
Step 3
Exam Tip
\(\alpha+\beta=6\) और \(\alpha^2+\beta^2=26\) है। \(36-2\alpha\beta=26\) से \(\alpha\beta=5\), इसलिए जड़ें (1) और (5) हैं।
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यदि \(x^2-7x+12=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(2\alpha+3\) और \(2\beta+3\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-7x+12=0\), which equation has roots \(2\alpha+3\) and \(2\beta+3\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-20x+99=0\)
B \(x^2-14x+99=0\)
C \(x^2-20x+91=0\)
D \(x^2+20x+99=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-20x+99=0\)
Step 1
Concept
The original roots are (3) and (4). The new roots are (9) and (11), so the equation is \(x^2-20x+99=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-20x+99=0\). The original roots are (3) and (4). The new roots are (9) and (11), so the equation is \(x^2-20x+99=0\).
Step 3
Exam Tip
मूल जड़ें (3) और (4) हैं। नई जड़ें (9) और (11) हैं, इसलिए समीकरण \(x^2-20x+99=0\) है।
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यदि \(x^2-5x+6=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha+\beta\) और \(\alpha\beta\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are roots of \(x^2-5x+6=0\), which equation has roots \(\alpha+\beta\) and \(\alpha\beta\)?
#quadratic-roots
#new-equation
#sum-product-roots
A \(x^2-11x+30=0\)
B \(x^2+11x+30=0\)
C \(x^2-5x+6=0\)
D \(x^2-30x+11=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-11x+30=0\)
Step 1
Concept
Here \(\alpha+\beta=5\) and \(\alpha\beta=6\). The new roots are (5) and (6), so the equation is \(x^2-11x+30=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-11x+30=0\). Here \(\alpha+\beta=5\) and \(\alpha\beta=6\). The new roots are (5) and (6), so the equation is \(x^2-11x+30=0\).
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha\beta=6\) हैं। नई जड़ें (5) और (6) हैं, इसलिए समीकरण \(x^2-11x+30=0\) है।
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यदि \(x^2-5x+c=0\) की जड़ें \(\alpha,\beta\) हैं और \(\alpha^2+\beta^2=17\), तो जड़ें क्या हैं?
If \(\alpha,\beta\) are roots of \(x^2-5x+c=0\) and \(\alpha^2+\beta^2=17\), what are the roots?
#quadratic-roots
#determine-roots
#sum-product
A (1) और (4) / (1) and (4)
B (2) और (3) / (2) and (3)
C (0) और (5) / (0) and (5)
D (-1) और (6) / (-1) and (6)
Explanation opens after your attempt
Correct Answer
A. (1) और (4) / (1) and (4)
Step 1
Concept
Here \(\alpha+\beta=5\) and \(\alpha^2+\beta^2=17\). From \(25-2\alpha\beta=17\), \(\alpha\beta=4\), so the roots are (1) and (4).
Step 2
Why this answer is correct
The correct answer is A. (1) और (4) / (1) and (4). Here \(\alpha+\beta=5\) and \(\alpha^2+\beta^2=17\). From \(25-2\alpha\beta=17\), \(\alpha\beta=4\), so the roots are (1) and (4).
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha^2+\beta^2=17\) है। \(25-2\alpha\beta=17\) से \(\alpha\beta=4\), इसलिए जड़ें (1) और (4) हैं।
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यदि \(x^2-9x+14=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha-3\) और \(\beta-3\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-9x+14=0\), which equation has roots \(\alpha-3\) and \(\beta-3\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-3x-4=0\)
B \(x^2+3x-4=0\)
C \(x^2-3x+4=0\)
D \(x^2-9x+14=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3x-4=0\)
Step 1
Concept
The original roots are (2) and (7). The new roots are (-1) and (4), so the equation is \(x^2-3x-4=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x-4=0\). The original roots are (2) and (7). The new roots are (-1) and (4), so the equation is \(x^2-3x-4=0\).
Step 3
Exam Tip
मूल जड़ें (2) और (7) हैं। नई जड़ें (-1) और (4) होंगी, इसलिए समीकरण \(x^2-3x-4=0\) है।
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यदि \(x^2-4x+c=0\) की जड़ें \(\alpha,\beta\) हैं और \(\alpha^2+\beta^2=10\), तो समीकरण की जड़ें क्या हैं?
If \(\alpha,\beta\) are roots of \(x^2-4x+c=0\) and \(\alpha^2+\beta^2=10\), what are the roots of the equation?
#quadratic-roots
#determine-roots
#sum-product
A (1) और (3) / (1) and (3)
B (2) और (2) / (2) and (2)
C (0) और (4) / (0) and (4)
D (-1) और (5) / (-1) and (5)
Explanation opens after your attempt
Correct Answer
A. (1) और (3) / (1) and (3)
Step 1
Concept
Here \(\alpha+\beta=4\) and \(\alpha^2+\beta^2=10\). From \(16-2\alpha\beta=10\), \(\alpha\beta=3\), so the roots are (1) and (3).
Step 2
Why this answer is correct
The correct answer is A. (1) और (3) / (1) and (3). Here \(\alpha+\beta=4\) and \(\alpha^2+\beta^2=10\). From \(16-2\alpha\beta=10\), \(\alpha\beta=3\), so the roots are (1) and (3).
Step 3
Exam Tip
\(\alpha+\beta=4\) और \(\alpha^2+\beta^2=10\) है। \(16-2\alpha\beta=10\) से \(\alpha\beta=3\), इसलिए जड़ें (1) और (3) हैं।
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यदि \(x^2-6x+5=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(3\alpha-2\) और \(3\beta-2\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-6x+5=0\), which equation has roots \(3\alpha-2\) and \(3\beta-2\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-14x+13=0\)
B \(x^2-18x+45=0\)
C \(x^2-14x+25=0\)
D \(x^2+14x+13=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-14x+13=0\)
Step 1
Concept
The original roots are (1) and (5), so the new roots are (1) and (13). Their equation is \(x^2-14x+13=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-14x+13=0\). The original roots are (1) and (5), so the new roots are (1) and (13). Their equation is \(x^2-14x+13=0\).
Step 3
Exam Tip
मूल जड़ें (1) और (5) हैं, इसलिए नई जड़ें (1) और (13) हैं। उनका समीकरण \(x^2-14x+13=0\) है।
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यदि \(x^2+px+q=0\) की जड़ें \(\alpha,\beta\) हैं और \(\alpha+1,\beta+1\), \(x^2-5x+6=0\) की जड़ें हैं, तो (p,q) क्या होंगे?
If \(\alpha,\beta\) are the roots of \(x^2+px+q=0\) and \(\alpha+1,\beta+1\) are the roots of \(x^2-5x+6=0\), what are (p,q)?
#quadratic-roots
#transformed-roots
#parameter
A (p=-3,\ q=2)
B (p=3,\ q=2)
C (p=-2,\ q=3)
D (p=2,\ q=-3)
Explanation opens after your attempt
Correct Answer
A. (p=-3,\ q=2)
Step 1
Concept
The sum of new roots is \(\alpha+\beta+2=5\), so (p=-3). From product (q-p+1=6), we get (q=2).
Step 2
Why this answer is correct
The correct answer is A. (p=-3,\ q=2). The sum of new roots is \(\alpha+\beta+2=5\), so (p=-3). From product (q-p+1=6), we get (q=2).
Step 3
Exam Tip
नई जड़ों का योग \(\alpha+\beta+2=5\) है, इसलिए (p=-3)। गुणनफल (q-p+1=6) से (q=2) मिलता है।
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यदि \(x^2-3x-2=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha^2,\beta^2\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-3x-2=0\), which equation has roots \(\alpha^2,\beta^2\)?
#quadratic-roots
#squared-roots
#new-equation
A \(x^2-13x+4=0\)
B \(x^2+13x+4=0\)
C \(x^2-9x+4=0\)
D \(x^2-13x-4=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-13x+4=0\)
Step 1
Concept
Here \(\alpha+\beta=3\) and \(\alpha\beta=-2\). Thus \(\alpha^2+\beta^2=13\) and \(\alpha^2\beta^2=4\), so the equation is \(x^2-13x+4=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-13x+4=0\). Here \(\alpha+\beta=3\) and \(\alpha\beta=-2\). Thus \(\alpha^2+\beta^2=13\) and \(\alpha^2\beta^2=4\), so the equation is \(x^2-13x+4=0\).
Step 3
Exam Tip
\(\alpha+\beta=3\) और \(\alpha\beta=-2\) है। इसलिए \(\alpha^2+\beta^2=13\) और \(\alpha^2\beta^2=4\), अतः समीकरण \(x^2-13x+4=0\) है।
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\(x^2-5x+6=0\) की जड़ें \(\alpha,\beta\) हैं। \(\alpha+1,\beta+1\) जड़ों वाला समीकरण कौन-सा है?
The roots of \(x^2-5x+6=0\) are \(\alpha,\beta\). Which equation has roots \(\alpha+1,\beta+1\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-7x+12=0\)
B \(x^2-5x+12=0\)
C \(x^2-6x+8=0\)
D \(x^2+7x+12=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+12=0\)
Step 1
Concept
The original roots are (2) and (3), so the new roots are (3) and (4). Their equation is \(x^2-7x+12=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-7x+12=0\). The original roots are (2) and (3), so the new roots are (3) and (4). Their equation is \(x^2-7x+12=0\).
Step 3
Exam Tip
मूल जड़ें (2) और (3) हैं, इसलिए नई जड़ें (3) और (4) होंगी। उनका समीकरण \(x^2-7x+12=0\) है।
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यदि \(3x^2-10x+3=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{1}{\alpha},\frac{1}{\beta}\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(3x^2-10x+3=0\), which equation has roots \(\frac{1}{\alpha},\frac{1}{\beta}\)?
#quadratic-roots
#reciprocal-roots
#new-equation
A \(3x^2-10x+3=0\)
B \(3x^2+10x+3=0\)
C \(x^2-10x+3=0\)
D \(10x^2-3x+3=0\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2-10x+3=0\)
Step 1
Concept
Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=1\). The reciprocal roots also have sum \(\frac{10}{3}\) and product (1).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-10x+3=0\). Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=1\). The reciprocal roots also have sum \(\frac{10}{3}\) and product (1).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{10}{3}\) और \(\alpha\beta=1\) है। व्युत्क्रम जड़ों का योग \(\frac{10}{3}\) और गुणनफल (1) ही रहता है।
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यदि \(D_1=64\), \(D_2=15\), \(D_3=0\) और \(D_4=-9\) हों, तो अपरिमेय असमान मूल किसमें होंगे?
If \(D_1=64\), \(D_2=15\), \(D_3=0\), and \(D_4=-9\), which one gives irrational distinct roots?
#quadratic-equations
#concept-check
#irrational-roots
A \(D_2=15\)
B \(D_1=64\)
C \(D_3=0\)
D \(D_4=-9\)
Explanation opens after your attempt
Correct Answer
A. \(D_2=15\)
Step 1
Concept
For irrational distinct roots, (D>0) and (D) must not be a perfect square. (15) satisfies this condition.
Step 2
Why this answer is correct
The correct answer is A. \(D_2=15\). For irrational distinct roots, (D>0) and (D) must not be a perfect square. (15) satisfies this condition.
Step 3
Exam Tip
अपरिमेय असमान मूलों के लिए (D>0) और (D) पूर्ण वर्ग नहीं होना चाहिए। (15) यह शर्त पूरी करता है।
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यदि \(D_1=36\), \(D_2=11\), \(D_3=0\) और \(D_4=-5\) हों, तो अपरिमेय असमान मूल किसमें होंगे?
If \(D_1=36\), \(D_2=11\), \(D_3=0\), and \(D_4=-5\), which one gives irrational distinct roots?
#quadratic-equations
#concept-check
#irrational-roots
A \(D_2=11\)
B \(D_1=36\)
C \(D_3=0\)
D \(D_4=-5\)
Explanation opens after your attempt
Correct Answer
A. \(D_2=11\)
Step 1
Concept
For irrational distinct roots, (D>0) and (D) must not be a perfect square. (11) satisfies this condition.
Step 2
Why this answer is correct
The correct answer is A. \(D_2=11\). For irrational distinct roots, (D>0) and (D) must not be a perfect square. (11) satisfies this condition.
Step 3
Exam Tip
अपरिमेय असमान मूलों के लिए (D>0) और (D) पूर्ण वर्ग नहीं होना चाहिए। (11) इसी शर्त को पूरा करता है।
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यदि दो वास्तविक मूलों का गुणनफल धनात्मक और योग धनात्मक है तो दोनों मूल कैसे होंगे?
If the product of two real roots is positive and their sum is positive, how will both roots be?
#roots
#sign_of_roots
#reasoning
A दोनों धनात्मक / Both positive
B दोनों ऋणात्मक / Both negative
C एक धनात्मक और एक ऋणात्मक / One positive and one negative
D एक मूल (0) / One root is (0)
Explanation opens after your attempt
Correct Answer
A. दोनों धनात्मक / Both positive
Step 1
Concept
A positive product means both signs are same. A positive sum means both roots are positive.
Step 2
Why this answer is correct
The correct answer is A. दोनों धनात्मक / Both positive. A positive product means both signs are same. A positive sum means both roots are positive.
Step 3
Exam Tip
गुणनफल धनात्मक होने पर दोनों चिन्ह समान होते हैं। योग धनात्मक होने से दोनों मूल धनात्मक होंगे।
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यदि दो वास्तविक मूलों का गुणनफल धनात्मक और योग ऋणात्मक है तो दोनों मूल कैसे होंगे?
If the product of two real roots is positive and their sum is negative, how will both roots be?
#roots
#sign_of_roots
#reasoning
A दोनों धनात्मक / Both positive
B दोनों ऋणात्मक / Both negative
C एक धनात्मक और एक ऋणात्मक / One positive and one negative
D एक मूल (0) / One root is (0)
Explanation opens after your attempt
Correct Answer
B. दोनों ऋणात्मक / Both negative
Step 1
Concept
A positive product means both roots have the same sign. A negative sum means both roots are negative.
Step 2
Why this answer is correct
The correct answer is B. दोनों ऋणात्मक / Both negative. A positive product means both roots have the same sign. A negative sum means both roots are negative.
Step 3
Exam Tip
गुणनफल धनात्मक होने पर दोनों मूलों का चिन्ह समान होता है। योग ऋणात्मक होने से दोनों मूल ऋणात्मक होंगे।
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यदि (D_1=(y+5)2 ), (D_2=-(y+5)2 ) और \(y\neq-5\), तो दो असमान वास्तविक मूल किस स्थिति में होंगे?
If (D_1=(y+5)2 ), (D_2=-(y+5)2 ), and \(y\neq-5\), which case gives two distinct real roots?
#quadratic-equations
#concept-check
#discriminant
A (D_1=(y+5)2 )
B (D_2=-(y+5)2 )
C दोनों / Both
D कोई नहीं / None
Explanation opens after your attempt
Correct Answer
A. (D_1=(y+5)2 )
Step 1
Concept
Since \(y\neq-5\), ((y+5)2 >0). A positive (D) gives two distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. (D_1=(y+5)2 ). Since \(y\neq-5\), ((y+5)2 >0). A positive (D) gives two distinct real roots.
Step 3
Exam Tip
क्योंकि \(y\neq-5\), इसलिए ((y+5)2 >0) है। धनात्मक (D) दो असमान वास्तविक मूल देता है।
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कथन: (x-2 -2(a-3b)x+(a+3b)2 =0) में यदि (ab<0), तो मूल वास्तविक और असमान हैं। कारण: इसका (D=-48ab) है। सही विकल्प चुनिए।
Assertion: In (x-2 -2(a-3b)x+(a+3b)2 =0), if (ab<0), the roots are real and distinct. Reason: Its (D=-48ab). Choose the correct option.
#quadratic-equations
#assertion-reason
#parameter
A कथन और कारण दोनों सही हैं / Both assertion and reason are correct
B कथन सही है, कारण गलत है / Assertion is correct, reason is wrong
C कथन गलत है, कारण सही है / Assertion is wrong, reason is correct
D कथन और कारण दोनों गलत हैं / Both assertion and reason are wrong
Explanation opens after your attempt
Correct Answer
A. कथन और कारण दोनों सही हैं / Both assertion and reason are correct
Step 1
Concept
The discriminant is (D=-48ab). If (ab<0), then (D>0), so roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. कथन और कारण दोनों सही हैं / Both assertion and reason are correct. The discriminant is (D=-48ab). If (ab<0), then (D>0), so roots are real and distinct.
Step 3
Exam Tip
विविक्तकर (D=-48ab) है। (ab<0) होने पर (D>0), इसलिए मूल वास्तविक असमान होंगे।
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यदि (D_1=(v-3)2 ), (D_2=-(v-3)2 ) और \(v\neq3\), तो दो असमान वास्तविक मूल किस स्थिति में होंगे?
If (D_1=(v-3)2 ), (D_2=-(v-3)2 ), and \(v\neq3\), which case gives two distinct real roots?
#quadratic-equations
#concept-check
#discriminant
A (D_1=(v-3)2 )
B (D_2=-(v-3)2 )
C दोनों / Both
D कोई नहीं / None
Explanation opens after your attempt
Correct Answer
A. (D_1=(v-3)2 )
Step 1
Concept
Since \(v\neq3\), ((v-3)2 >0). A positive discriminant gives two distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. (D_1=(v-3)2 ). Since \(v\neq3\), ((v-3)2 >0). A positive discriminant gives two distinct real roots.
Step 3
Exam Tip
क्योंकि \(v\neq3\), इसलिए ((v-3)2 >0) है। धनात्मक विविक्तकर दो असमान वास्तविक मूल देता है।
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कथन: (x-2 -2(a-b)x+(a+b)2 =0) में यदि (ab<0), तो मूल वास्तविक और असमान हैं। कारण: इसका (D=-16ab) है। सही विकल्प चुनिए।
Assertion: In (x-2 -2(a-b)x+(a+b)2 =0), if (ab<0), the roots are real and distinct. Reason: Its (D=-16ab). Choose the correct option.
#quadratic-equations
#assertion-reason
#parameter
A कथन और कारण दोनों सही हैं / Both assertion and reason are correct
B कथन सही है, कारण गलत है / Assertion is correct, reason is wrong
C कथन गलत है, कारण सही है / Assertion is wrong, reason is correct
D कथन और कारण दोनों गलत हैं / Both assertion and reason are wrong
Explanation opens after your attempt
Correct Answer
A. कथन और कारण दोनों सही हैं / Both assertion and reason are correct
Step 1
Concept
The discriminant is (D=-16ab). If (ab<0), then (D>0), so roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. कथन और कारण दोनों सही हैं / Both assertion and reason are correct. The discriminant is (D=-16ab). If (ab<0), then (D>0), so roots are real and distinct.
Step 3
Exam Tip
विविक्तकर (D=-16ab) है। यदि (ab<0), तो (D>0) और मूल वास्तविक असमान होंगे।
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यदि \(D_1=4r^2\), \(D_2=-r^2\), \(D_3=0\) और \(r\neq0\), तो दो असमान वास्तविक मूल किसमें होंगे?
If \(D_1=4r^2\), \(D_2=-r^2\), \(D_3=0\), and \(r\neq0\), which gives two distinct real roots?
#quadratic-equations
#concept-check
#discriminant
A \(D_1=4r^2\)
B \(D_2=-r^2\)
C \(D_3=0\)
D \(D_2\) और \(D_3\) / \(D_2\) and \(D_3\)
Explanation opens after your attempt
Correct Answer
A. \(D_1=4r^2\)
Step 1
Concept
Since \(r\neq0\), \(4r^2>0\). A positive (D) gives two distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. \(D_1=4r^2\). Since \(r\neq0\), \(4r^2>0\). A positive (D) gives two distinct real roots.
Step 3
Exam Tip
क्योंकि \(r\neq0\), इसलिए \(4r^2>0\) है। धनात्मक (D) दो असमान वास्तविक मूल देता है।
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कथन: (x-2 -2(a+b)x+(a-b)2 =0) में यदि (ab>0), तो मूल वास्तविक और असमान हैं। कारण: इसका (D=16ab) है। सही विकल्प चुनिए।
Assertion: In (x-2 -2(a+b)x+(a-b)2 =0), if (ab>0), the roots are real and distinct. Reason: Its (D=16ab). Choose the correct option.
#quadratic-equations
#assertion-reason
#parameter
A कथन और कारण दोनों सही हैं / Both assertion and reason are correct
B कथन सही है, कारण गलत है / Assertion is correct, reason is wrong
C कथन गलत है, कारण सही है / Assertion is wrong, reason is correct
D कथन और कारण दोनों गलत हैं / Both assertion and reason are wrong
Explanation opens after your attempt
Correct Answer
A. कथन और कारण दोनों सही हैं / Both assertion and reason are correct
Step 1
Concept
The discriminant (D=16ab) is correct. If (ab>0), then (D>0), so the roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. कथन और कारण दोनों सही हैं / Both assertion and reason are correct. The discriminant (D=16ab) is correct. If (ab>0), then (D>0), so the roots are real and distinct.
Step 3
Exam Tip
विविक्तकर (D=16ab) सही है। (ab>0) होने पर (D>0), इसलिए मूल वास्तविक और असमान होंगे।
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यदि \(2x^2-4x+n=0\) के मूल वास्तविक और भिन्न हैं, तो (n) के लिए कौन सी शर्त सही है?
If \(2x^2-4x+n=0\) has real and distinct roots, which condition is correct for (n)?
#quadratic_equations
#nature_of_roots
#inequality
A (n<2)
B (n=2)
C (n>2)
D \(n\le2\)
Explanation opens after your attempt
Step 1
Concept
For real and distinct roots, (D>0), so (16-8n>0) gives (n<2). In exams, do not forget coefficient (a) in (4ac).
Step 2
Why this answer is correct
The correct answer is A. (n<2). For real and distinct roots, (D>0), so (16-8n>0) gives (n<2). In exams, do not forget coefficient (a) in (4ac).
Step 3
Exam Tip
दो भिन्न वास्तविक मूलों के लिए (D>0), अतः (16-8n>0) से (n<2)। परीक्षा में coefficient (a) को (4ac) में शामिल करना न भूलें।
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किस (r) के लिए \(rx^2-5x+r=0\) के दो भिन्न वास्तविक मूल होंगे?
For which (r) will \(rx^2-5x+r=0\) have two distinct real roots?
#quadratic_equations
#nature_of_roots
#parameter_inequality
A \(r^2<\frac{25}{4}\) और \(r\neq0\) / \(r^2<\frac{25}{4}\) and \(r\neq0\)
B \(r^2=\frac{25}{4}\)
C \(r^2>\frac{25}{4}\)
D (r=0)
Explanation opens after your attempt
Correct Answer
A. \(r^2<\frac{25}{4}\) और \(r\neq0\) / \(r^2<\frac{25}{4}\) and \(r\neq0\)
Step 1
Concept
For two distinct real roots, (D>0), so \(25-4r^2>0\). Also \(r\neq0\) is needed because the equation must remain quadratic.
Step 2
Why this answer is correct
The correct answer is A. \(r^2<\frac{25}{4}\) और \(r\neq0\) / \(r^2<\frac{25}{4}\) and \(r\neq0\). For two distinct real roots, (D>0), so \(25-4r^2>0\). Also \(r\neq0\) is needed because the equation must remain quadratic.
Step 3
Exam Tip
दो भिन्न वास्तविक मूलों के लिए (D>0), इसलिए \(25-4r^2>0\)। साथ में \(r\neq0\) चाहिए क्योंकि समीकरण द्विघात होना चाहिए।
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यदि \(x^2+5x+k=0\) के दो भिन्न वास्तविक मूल हैं, तो (k) के लिए सही शर्त क्या है?
If \(x^2+5x+k=0\) has two distinct real roots, what is the correct condition for (k)?
#quadratic_equations
#nature_of_roots
#inequality
A \(k<\frac{25}{4}\)
B \(k=\frac{25}{4}\)
C \(k>\frac{25}{4}\)
D \(k<-\frac{25}{4}\)
Explanation opens after your attempt
Correct Answer
A. \(k<\frac{25}{4}\)
Step 1
Concept
For two distinct real roots, (D>0), so (25-4k>0) gives \(k<\frac{25}{4}\). In exams, keep the inequality sign correct while solving.
Step 2
Why this answer is correct
The correct answer is A. \(k<\frac{25}{4}\). For two distinct real roots, (D>0), so (25-4k>0) gives \(k<\frac{25}{4}\). In exams, keep the inequality sign correct while solving.
Step 3
Exam Tip
दो भिन्न वास्तविक मूलों के लिए (D>0), इसलिए (25-4k>0) से \(k<\frac{25}{4}\)। परीक्षा में असमानता हल करते समय चिन्ह सही रखें।
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किस शर्त पर \(ax^2+bx+c=0\) के दो भिन्न वास्तविक मूल होंगे?
Under which condition will \(ax^2+bx+c=0\) have two distinct real roots?
#quadratic_equations
#nature_of_roots
#concept
A \(b^2-4ac>0\)
B \(b^2-4ac=0\)
C \(b^2-4ac<0\)
D \(b^2+4ac=0\)
Explanation opens after your attempt
Correct Answer
A. \(b^2-4ac>0\)
Step 1
Concept
For two distinct real roots, the discriminant \(D=b^2-4ac\) is positive. In exams, first calculate (D) to decide the nature.
Step 2
Why this answer is correct
The correct answer is A. \(b^2-4ac>0\). For two distinct real roots, the discriminant \(D=b^2-4ac\) is positive. In exams, first calculate (D) to decide the nature.
Step 3
Exam Tip
दो भिन्न वास्तविक मूलों के लिए विविक्तकर \(D=b^2-4ac\) धनात्मक होता है। परीक्षा में प्रकृति तय करने के लिए पहले (D) निकालें।
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यदि (2x-2 -2\(\mu+3\)x+\mu-2 +6\mu+5=0) के मूल वास्तविक और भिन्न हैं, तो \(\mu\) पर क्या शर्त है?
If (2x-2 -2\(\mu+3\)x+\mu-2 +6\mu+5=0) has real and distinct roots, what is the condition on \(\mu\)?
#quadratic equations
#parameter mu
#critical check
A सभी वास्तविक \(\mu\) / All real \(\mu\)
B कोई वास्तविक \(\mu\) नहीं / No real \(\mu\)
C \(\mu>0\)
D \(\mu<0\)
Explanation opens after your attempt
Correct Answer
A. सभी वास्तविक \(\mu\) / All real \(\mu\)
Step 1
Concept
Here (D=4\(\mu+3\)2 -8\(\mu^2+6\mu+5\)=-4\(\mu^2+6\mu+1\)). It is not always positive, so all \(\mu\) is not correct.
Step 2
Why this answer is correct
The correct answer is A. सभी वास्तविक \(\mu\) / All real \(\mu\). Here (D=4\(\mu+3\)2 -8\(\mu^2+6\mu+5\)=-4\(\mu^2+6\mu+1\)). It is not always positive, so all \(\mu\) is not correct.
Step 3
Exam Tip
यहाँ (D=4\(\mu+3\)2 -8\(\mu^2+6\mu+5\)=-4\(\mu^2+6\mu+1\)) है। यह हमेशा धनात्मक नहीं है, इसलिए सभी \(\mu\) सही नहीं है।
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यदि (D=(s-2 )(s+5)) है, तो मूल वास्तविक और भिन्न कब होंगे?
If (D=(s-2 )(s+5)), when will the roots be real and distinct?
#quadratic equations
#discriminant expression
#interval
A (s<-5) या (s>2) / (s<-5) or (s>2)
B (-5<s<2)
C (s=-5) या (s=2) / (s=-5) or (s=2)
D सभी वास्तविक (s) / All real (s)
Explanation opens after your attempt
Correct Answer
A. (s<-5) या (s>2) / (s<-5) or (s>2)
Step 1
Concept
For real and distinct roots, (D>0) is required. From ((s-2 )(s+5)>0), we get (s<-5) or (s>2).
Step 2
Why this answer is correct
The correct answer is A. (s<-5) या (s>2) / (s<-5) or (s>2). For real and distinct roots, (D>0) is required. From ((s-2 )(s+5)>0), we get (s<-5) or (s>2).
Step 3
Exam Tip
वास्तविक और भिन्न मूलों के लिए (D>0) चाहिए। ((s-2 )(s+5)>0) से (s<-5) या (s>2) मिलता है।
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यदि \(x^2-2px+p^2-5p=0\) के मूल वास्तविक और भिन्न हैं, तो (p) पर सही शर्त क्या है?
If \(x^2-2px+p^2-5p=0\) has real and distinct roots, what is the correct condition on (p)?
#quadratic equations
#parameter inequality
#D positive
A (p>0)
B (p<0)
C (p=0)
D \(p\ge0\)
Explanation opens after your attempt
Step 1
Concept
Here (D=4p-2 -4\(p^2-5p\)=20p). For real and distinct roots (D>0), hence (p>0).
Step 2
Why this answer is correct
The correct answer is A. (p>0). Here (D=4p-2 -4\(p^2-5p\)=20p). For real and distinct roots (D>0), hence (p>0).
Step 3
Exam Tip
यहाँ (D=4p-2 -4\(p^2-5p\)=20p) है। वास्तविक और भिन्न मूलों के लिए (D>0), अतः (p>0)।
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समीकरण \(x^2-6x+k=0\) के दो वास्तविक और असमान मूलों के लिए (k) पर कौन सी शर्त होगी?
What condition on (k) gives two real and distinct roots for \(x^2-6x+k=0\)?
#quadratic-equations
#inequality
#parameter
A (k<9)
B (k=9)
C (k>9)
D (k=0) मात्र / Only (k=0)
Explanation opens after your attempt
Step 1
Concept
Here (D=36-4k), and distinct real roots need (D>0). Hence (k<9).
Step 2
Why this answer is correct
The correct answer is A. (k<9). Here (D=36-4k), and distinct real roots need (D>0). Hence (k<9).
Step 3
Exam Tip
यहाँ (D=36-4k) है और असमान वास्तविक मूलों के लिए (D>0) चाहिए। इसलिए (k<9)।
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कथन: (D<0) होने पर दो वास्तविक और भिन्न मूल मिलते हैं। यह कथन कैसा है?
Statement: When (D<0), two real and distinct roots are obtained. What type of statement is this?
#quadratic equations
#true false
#D negative
A असत्य / False
B सत्य / True
C केवल पूर्ण वर्ग (D) पर सत्य / True only for perfect square (D)
D केवल (b=0) पर सत्य / True only when (b=0)
Explanation opens after your attempt
Correct Answer
A. असत्य / False
Step 1
Concept
When (D<0), real roots do not exist. Two real and distinct roots occur when (D>0).
Step 2
Why this answer is correct
The correct answer is A. असत्य / False. When (D<0), real roots do not exist. Two real and distinct roots occur when (D>0).
Step 3
Exam Tip
(D<0) होने पर वास्तविक मूल नहीं होते हैं। दो वास्तविक और भिन्न मूल (D>0) पर मिलते हैं।
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कौन सा समीकरण वास्तविक और भिन्न मूल रखता है?
Which equation has real and distinct roots?
#quadratic equations
#choose equation
#D positive
A \(x^2-3x+2=0\)
B \(x^2+2x+1=0\)
C \(x^2+x+1=0\)
D \(4x^2+4x+1=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3x+2=0\)
Step 1
Concept
In the first equation (D=(-3)2 -4(1)(2)=1>0). Hence its roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x+2=0\). In the first equation (D=(-3)2 -4(1)(2)=1>0). Hence its roots are real and distinct.
Step 3
Exam Tip
पहले समीकरण में (D=(-3)2 -4(1)(2)=1>0) है। इसलिए उसके मूल वास्तविक और भिन्न हैं।
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समीकरण \(3x^2+10x+3=0\) के मूलों की प्रकृति क्या है?
What is the nature of roots of \(3x^2+10x+3=0\)?
#quadratic equations
#nature of roots
#discriminant
#distinct roots
A दो वास्तविक और असमान / two real and distinct
B दो वास्तविक और समान / two real and equal
C वास्तविक मूल नहीं / no real roots
D दोनों मूल (0) / both roots are (0)
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और असमान / two real and distinct
Step 1
Concept
(D=102 -4(3)(3)=64>0). Therefore the roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और असमान / two real and distinct. (D=102 -4(3)(3)=64>0). Therefore the roots are real and distinct.
Step 3
Exam Tip
(D=102 -4(3)(3)=64>0) है। इसलिए मूल वास्तविक और असमान हैं।
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यदि (a<0), (c>0) और (b) कोई वास्तविक संख्या हो, तो \(ax^2+bx+c=0\) के मूलों की प्रकृति क्या होगी?
If (a<0), (c>0), and (b) is any real number, what will be the nature of roots of \(ax^2+bx+c=0\)?
#quadratic-equations
#sign-analysis
#distinct-roots
A दो वास्तविक और असमान / Two real and distinct
B दो वास्तविक और समान / Two real and equal
C कोई वास्तविक मूल नहीं / No real roots
D मूल हमेशा अपरिमेय / Roots are always irrational
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और असमान / Two real and distinct
Step 1
Concept
Here (ac<0), so (-4ac>0) and \(D=b^2-4ac>0\). Hence there will be two distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और असमान / Two real and distinct. Here (ac<0), so (-4ac>0) and \(D=b^2-4ac>0\). Hence there will be two distinct real roots.
Step 3
Exam Tip
यहाँ (ac<0), इसलिए (-4ac>0) और \(D=b^2-4ac>0\) है। अतः दो असमान वास्तविक मूल होंगे।
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यदि (a>0), (c<0) और (b) कोई वास्तविक संख्या हो, तो \(ax^2+bx+c=0\) के मूलों की प्रकृति क्या होगी?
If (a>0), (c<0), and (b) is any real number, what will be the nature of roots of \(ax^2+bx+c=0\)?
#quadratic-equations
#sign-analysis
#distinct-roots
A दो वास्तविक और असमान / Two real and distinct
B दो वास्तविक और समान / Two real and equal
C कोई वास्तविक मूल नहीं / No real roots
D मूल हमेशा अपरिमेय / Roots are always irrational
Explanation opens after your attempt
Correct Answer
A. दो वास्तविक और असमान / Two real and distinct
Step 1
Concept
Here (ac<0), so (-4ac>0) and \(D=b^2-4ac>0\). Hence there will be two distinct real roots.
Step 2
Why this answer is correct
The correct answer is A. दो वास्तविक और असमान / Two real and distinct. Here (ac<0), so (-4ac>0) and \(D=b^2-4ac>0\). Hence there will be two distinct real roots.
Step 3
Exam Tip
यहाँ (ac<0), इसलिए (-4ac>0) और \(D=b^2-4ac>0\) है। अतः दो असमान वास्तविक मूल होंगे।
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समीकरण \(x^2-6x+13=0\) के वास्तविक मूलों के बारे में क्या सही है?
What is correct about the real roots of the equation \(x^2-6x+13=0\)?
#quadratic equations
#nature of roots
#no real roots
A कोई वास्तविक मूल नहीं / No real roots
B दो वास्तविक और भिन्न मूल / Two real and distinct roots
C दो वास्तविक और समान मूल / Two real and equal roots
D एक वास्तविक मूल / One real root
Explanation opens after your attempt
Correct Answer
A. कोई वास्तविक मूल नहीं / No real roots
Step 1
Concept
Here (D=(-6)2 -4(1)(13)=-16<0). Therefore there is no real root.
Step 2
Why this answer is correct
The correct answer is A. कोई वास्तविक मूल नहीं / No real roots. Here (D=(-6)2 -4(1)(13)=-16<0). Therefore there is no real root.
Step 3
Exam Tip
यहाँ (D=(-6)2 -4(1)(13)=-16<0) है। इसलिए कोई वास्तविक मूल नहीं है।
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यदि \(7x^2-6x+\lambda=0\) की जड़ें वास्तविक नहीं हैं, तो \(\lambda\) पर सही शर्त क्या है?
If the roots of \(7x^2-6x+\lambda=0\) are not real, what is the correct condition on \(\lambda\)?
#quadratic-roots
#non-real-roots
#discriminant
A \(\lambda>\frac{9}{7}\)
B \(\lambda<\frac{9}{7}\)
C \(\lambda=\frac{9}{7}\)
D \(\lambda\le\frac{9}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(\lambda>\frac{9}{7}\)
Step 1
Concept
For non-real roots, (D<0) is required. From \(36-28\lambda<0\), we get \(\lambda>\frac{9}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(\lambda>\frac{9}{7}\). For non-real roots, (D<0) is required. From \(36-28\lambda<0\), we get \(\lambda>\frac{9}{7}\).
Step 3
Exam Tip
वास्तविक नहीं होने के लिए (D<0) चाहिए। \(36-28\lambda<0\) से \(\lambda>\frac{9}{7}\) मिलता है।
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(9x-2 -6(a-1)x+a-2 -4a-5=0) की जड़ें वास्तविक हों, तो सही शर्त क्या है?
What is the correct condition for (9x-2 -6(a-1)x+a-2 -4a-5=0) to have real roots?
#quadratic-roots
#real-roots
#parameter-condition
A \(a\ge-3\)
B \(a\le-3\)
C (a>3)
D (a< -3)
Explanation opens after your attempt
Correct Answer
A. \(a\ge-3\)
Step 1
Concept
Here (D=36(a-1)2 -36\(a^2-4a-5\)=72(a+3)). For real roots, \(D\ge0\), so \(a\ge-3\).
Step 2
Why this answer is correct
The correct answer is A. \(a\ge-3\). Here (D=36(a-1)2 -36\(a^2-4a-5\)=72(a+3)). For real roots, \(D\ge0\), so \(a\ge-3\).
Step 3
Exam Tip
यहाँ (D=36(a-1)2 -36\(a^2-4a-5\)=72(a+3)) है। वास्तविक जड़ों के लिए \(D\ge0\), इसलिए \(a\ge-3\)।
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(9x-2 -6(a-1)x+a-2 -4a-5=0) की जड़ें वास्तविक हों, तो (a) पर सही शर्त क्या है?
For (9x-2 -6(a-1)x+a-2 -4a-5=0) to have real roots, what is the correct condition on (a)?
#quadratic-roots
#real-roots
#error-check
A \(a\ge-\frac{7}{2}\)
B \(a\le-\frac{7}{2}\)
C (a>1)
D (a<5)
Explanation opens after your attempt
Correct Answer
A. \(a\ge-\frac{7}{2}\)
Step 1
Concept
For real roots, \(D\ge0\) is required. Here (D=36(a-1)2 -36\(a^2-4a-5\)=72a+216), so the exact condition is \(a\ge-3\), not \(a\ge-\frac{7}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(a\ge-\frac{7}{2}\). For real roots, \(D\ge0\) is required. Here (D=36(a-1)2 -36\(a^2-4a-5\)=72a+216), so the exact condition is \(a\ge-3\), not \(a\ge-\frac{7}{2}\).
Step 3
Exam Tip
वास्तविक जड़ों के लिए \(D\ge0\) चाहिए। यहाँ (D=36(a-1)2 -36\(a^2-4a-5\)=72a+216), इसलिए \(a\ge-\frac{7}{2}\) नहीं बल्कि \(a\ge-3\) होगा।
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