100 results found for "always distinct roots" in Class 10.
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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\(x^2+12x+\lambda=0\) 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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\(x^2-10x+k=0\) 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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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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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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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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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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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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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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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 \(x^2+px+6=0\) 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 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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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 (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 (x-2 -2(a+b)x+2ab=0) के मूल वास्तविक हों, तो (a) और (b) के लिए कौन सा कथन हमेशा सही है?
If (x-2 -2(a+b)x+2ab=0) has real roots, which statement is always true for (a) and (b)?
#quadratic-equations
#algebraic-parameter
#real-roots
A \(a^2+b^2\geq0\) के कारण मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2+b^2\geq0\)
B मूल तभी वास्तविक हैं जब (ab>0) / Roots are real only when (ab>0)
C मूल कभी वास्तविक नहीं होते / Roots are never real
D मूल तभी समान हैं जब (a+b=0) / Roots are equal only when (a+b=0)
Explanation opens after your attempt
Correct Answer
A. \(a^2+b^2\geq0\) के कारण मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2+b^2\geq0\)
Step 1
Concept
Here (D=4(a+b)2 -8ab=4\(a^2+b^2\)). It is always zero or positive, so real roots exist.
Step 2
Why this answer is correct
The correct answer is A. \(a^2+b^2\geq0\) के कारण मूल हमेशा वास्तविक हैं / Roots are always real because \(a^2+b^2\geq0\). Here (D=4(a+b)2 -8ab=4\(a^2+b^2\)). It is always zero or positive, so real roots exist.
Step 3
Exam Tip
यहाँ (D=4(a+b)2 -8ab=4\(a^2+b^2\)) है। यह हमेशा (0) या धनात्मक होता है, इसलिए वास्तविक मूल मिलते हैं।
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समीकरण \(x^2+2px+p^2-1=0\) के मूलों की प्रकृति क्या है?
What is the nature of roots of \(x^2+2px+p^2-1=0\)?
#quadratic equations
#always distinct roots
#parameter
A हमेशा वास्तविक और भिन्न / Always real and distinct
B हमेशा वास्तविक और समान / Always real and equal
C हमेशा वास्तविक नहीं / Always not real
D केवल (p=1) पर वास्तविक / Real only when (p=1)
Explanation opens after your attempt
Correct Answer
A. हमेशा वास्तविक और भिन्न / Always real and distinct
Step 1
Concept
Here (D=(2p)2 -4\(p^2-1\)=4). So for every real (p), roots are real and distinct.
Step 2
Why this answer is correct
The correct answer is A. हमेशा वास्तविक और भिन्न / Always real and distinct. Here (D=(2p)2 -4\(p^2-1\)=4). So for every real (p), roots are real and distinct.
Step 3
Exam Tip
यहाँ (D=(2p)2 -4\(p^2-1\)=4) है। इसलिए हर वास्तविक (p) के लिए मूल वास्तविक और भिन्न हैं।
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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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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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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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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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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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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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(x-2 -2x+\(a^2+3\)=0) की जड़ों की प्रकृति क्या है?
What is the nature of the roots of (x-2 -2x+\(a^2+3\)=0)?
#quadratic-roots
#nature-of-roots
#always-non-real
A हर वास्तविक (a) के लिए वास्तविक नहीं / Not real for every real (a)
B हर वास्तविक (a) के लिए समान वास्तविक / Equal real for every real (a)
C हर वास्तविक (a) के लिए दो वास्तविक भिन्न / Two real distinct for every real (a)
D एक जड़ हमेशा (0) है / One root is always (0)
Explanation opens after your attempt
Correct Answer
A. हर वास्तविक (a) के लिए वास्तविक नहीं / Not real for every real (a)
Step 1
Concept
The discriminant is (D=4-4\(a^2+3\)=-4a-2 -8). It is negative for every real (a), so the roots are not real.
Step 2
Why this answer is correct
The correct answer is A. हर वास्तविक (a) के लिए वास्तविक नहीं / Not real for every real (a). The discriminant is (D=4-4\(a^2+3\)=-4a-2 -8). It is negative for every real (a), so the roots are not real.
Step 3
Exam Tip
विविक्तकर (D=4-4\(a^2+3\)=-4a-2 -8) है। यह हर वास्तविक (a) के लिए ऋणात्मक है, इसलिए जड़ें वास्तविक नहीं हैं।
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(x-2 -2x+\(a^2+2\)=0) की जड़ों की प्रकृति क्या है?
What is the nature of the roots of (x-2 -2x+\(a^2+2\)=0)?
#quadratic-roots
#nature-of-roots
#always-non-real
A दो वास्तविक भिन्न / Two real distinct
B समान वास्तविक / Equal real
C हर (a) के लिए वास्तविक नहीं / Not real for every (a)
D एक जड़ शून्य / One root is zero
Explanation opens after your attempt
Correct Answer
C. हर (a) के लिए वास्तविक नहीं / Not real for every (a)
Step 1
Concept
The discriminant is (D=4-4\(a^2+2\)=-4\(a^2+1\)). It is negative for every real (a), so the roots are not real.
Step 2
Why this answer is correct
The correct answer is C. हर (a) के लिए वास्तविक नहीं / Not real for every (a). The discriminant is (D=4-4\(a^2+2\)=-4\(a^2+1\)). It is negative for every real (a), so the roots are not real.
Step 3
Exam Tip
विविक्तकर (D=4-4\(a^2+2\)=-4\(a^2+1\)) है। यह हर वास्तविक (a) के लिए ऋणात्मक है, इसलिए जड़ें वास्तविक नहीं 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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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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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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कथन: (D>0) होने पर द्विघात समीकरण के मूल हमेशा बराबर होते हैं। यह कथन कैसा है?
Statement: When (D>0), the roots of a quadratic equation are always equal. How is this statement?
#quadratic_equations
#nature_of_roots
#common_mistake
A सत्य / True
B असत्य / False
C केवल (D=1) पर सत्य / True only when (D=1)
D केवल (a=1) पर सत्य / True only when (a=1)
Explanation opens after your attempt
Correct Answer
B. असत्य / False
Step 1
Concept
When (D>0), roots are not equal; they are distinct real roots. In exams, equal roots occur only when (D=0).
Step 2
Why this answer is correct
The correct answer is B. असत्य / False. When (D>0), roots are not equal; they are distinct real roots. In exams, equal roots occur only when (D=0).
Step 3
Exam Tip
(D>0) होने पर मूल बराबर नहीं, बल्कि भिन्न वास्तविक होते हैं। परीक्षा में बराबर मूल केवल (D=0) पर आते हैं।
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कथन: (D>0) होने पर मूल हमेशा समान होते हैं। सही विकल्प चुनिए।
Statement: When (D>0), roots are always equal. Choose the correct option.
#quadratic equations
#common mistake
#D positive
A कथन गलत है / The statement is wrong
B कथन सही है / The statement is correct
C कथन (D=0) पर लागू है / The statement applies to (D=0)
D कथन (D<0) पर लागू है / The statement applies to (D<0)
Explanation opens after your attempt
Correct Answer
A. कथन गलत है / The statement is wrong
Step 1
Concept
When (D>0), roots are real and distinct. Equal roots occur only when (D=0).
Step 2
Why this answer is correct
The correct answer is A. कथन गलत है / The statement is wrong. When (D>0), roots are real and distinct. Equal roots occur only when (D=0).
Step 3
Exam Tip
(D>0) पर मूल वास्तविक और भिन्न होते हैं। समान मूल केवल (D=0) पर होते हैं।
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किसी बहुपद के लिए (p(1)=0), (p(2)=0), (p(3)=0) है। यदि ये तीनों अलग शून्यक हैं, तो ग्राफ (x)-अक्ष से कितने अलग बिंदुओं पर मिलेगा?
For a polynomial (p(1)=0), (p(2)=0), (p(3)=0). If these are three distinct zeroes, at how many distinct points will the graph meet the (x)-axis?
#distinct zeroes
#graph
#count
A एक / One
B दो / Two
C तीन / Three
D छह / Six
Explanation opens after your attempt
Correct Answer
C. तीन / Three
Step 1
Concept
Three distinct (x)-values give three distinct (x)-axis points. Tip: distinct zeroes make distinct intersection points.
Step 2
Why this answer is correct
The correct answer is C. तीन / Three. Three distinct (x)-values give three distinct (x)-axis points. Tip: distinct zeroes make distinct intersection points.
Step 3
Exam Tip
तीन अलग (x)-मान तीन अलग (x)-अक्ष बिंदु देते हैं। टिप: अलग शून्यक अलग कटान बिंदु बनाते हैं।
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यदि (x-2 -2(k-3)x+k-2 -6k+8=0) है, तो मूलों की प्रकृति क्या है?
If (x-2 -2(k-3)x+k-2 -6k+8=0), what is the nature of roots?
#quadratic equations
#always real distinct
#constant D
A हमेशा वास्तविक और भिन्न / Always real and distinct
B हमेशा वास्तविक और समान / Always real and equal
C वास्तविक मूल नहीं / No real roots
D प्रकृति (k) पर निर्भर / Nature depends on (k)
Explanation opens after your attempt
Correct Answer
A. हमेशा वास्तविक और भिन्न / Always real and distinct
Step 1
Concept
Here (D=4(k-3)2 -4\(k^2-6k+8\)=4>0). Therefore roots are always real and distinct.
Step 2
Why this answer is correct
The correct answer is A. हमेशा वास्तविक और भिन्न / Always real and distinct. Here (D=4(k-3)2 -4\(k^2-6k+8\)=4>0). Therefore roots are always real and distinct.
Step 3
Exam Tip
यहाँ (D=4(k-3)2 -4\(k^2-6k+8\)=4>0) है। इसलिए मूल हमेशा वास्तविक और भिन्न हैं।
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यदि (D=(a-1)2 +5) है, तो मूलों की प्रकृति के बारे में सही कथन क्या है?
If (D=(a-1)2 +5), what is the correct statement about the nature of roots?
#quadratic equations
#positive D
#always distinct
A हमेशा वास्तविक और भिन्न / Always real and distinct
B हमेशा वास्तविक और समान / Always real and equal
C हमेशा वास्तविक नहीं / Always not real
D प्रकृति तय नहीं हो सकती / Nature cannot be decided
Explanation opens after your attempt
Correct Answer
A. हमेशा वास्तविक और भिन्न / Always real and distinct
Step 1
Concept
For every real (a), ((a-1)2 +5>0). Therefore the roots will always be real and distinct.
Step 2
Why this answer is correct
The correct answer is A. हमेशा वास्तविक और भिन्न / Always real and distinct. For every real (a), ((a-1)2 +5>0). Therefore the roots will always be real and distinct.
Step 3
Exam Tip
किसी भी वास्तविक (a) के लिए ((a-1)2 +5>0) है। इसलिए मूल हमेशा वास्तविक और भिन्न होंगे।
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समीकरण (2x-2 -4ax+\(2a^2-3\)=0) के मूलों की प्रकृति क्या है?
What is the nature of roots of (2x-2 -4ax+\(2a^2-3\)=0)?
#quadratic equations
#always real distinct
#constant D
A हमेशा वास्तविक और भिन्न / Always real and distinct
B हमेशा वास्तविक और समान / Always real and equal
C हमेशा वास्तविक नहीं / Always not real
D सिर्फ (a=0) पर वास्तविक / Real only when (a=0)
Explanation opens after your attempt
Correct Answer
A. हमेशा वास्तविक और भिन्न / Always real and distinct
Step 1
Concept
Here (D=(-4a)2 -4(2)\(2a^2-3\)=24>0). Thus roots are always real and distinct.
Step 2
Why this answer is correct
The correct answer is A. हमेशा वास्तविक और भिन्न / Always real and distinct. Here (D=(-4a)2 -4(2)\(2a^2-3\)=24>0). Thus roots are always real and distinct.
Step 3
Exam Tip
यहाँ (D=(-4a)2 -4(2)\(2a^2-3\)=24>0) है। इसलिए मूल हमेशा वास्तविक और भिन्न हैं।
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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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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 (x-2 -2(a+3)x+a-2 +6a+5=0) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha-\beta\) का धनात्मक मान क्या है?
If \(\alpha,\beta\) are the roots of (x-2 -2(a+3)x+a-2 +6a+5=0), what is the positive value of \(\alpha-\beta\)?
#quadratic-roots
#parametric-roots
#difference-of-roots
A (4)
B (2)
C (1)
D (3)
Explanation opens after your attempt
Step 1
Concept
The equation becomes ((x-(a+1))(x-(a+5))=0). So the roots are (a+1) and (a+5), hence the positive difference is (4).
Step 2
Why this answer is correct
The correct answer is A. (4). The equation becomes ((x-(a+1))(x-(a+5))=0). So the roots are (a+1) and (a+5), hence the positive difference is (4).
Step 3
Exam Tip
यहाँ समीकरण ((x-(a+1))(x-(a+5))=0) बनता है। इसलिए जड़ें (a+1) और (a+5) हैं, अतः धनात्मक अंतर (4) है।
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यदि \(x^2-12x+m=0\) की दोनों जड़ें अभाज्य संख्याएँ हैं, तो (m) का मान क्या है?
If both roots of \(x^2-12x+m=0\) are prime numbers, what is the value of (m)?
#quadratic-roots
#prime-roots
#integer-roots
A (30)
B (35)
C (40)
D (45)
Explanation opens after your attempt
Step 1
Concept
The prime roots with sum (12) are (5) and (7). Their product is (35), so (m=35).
Step 2
Why this answer is correct
The correct answer is B. (35). The prime roots with sum (12) are (5) and (7). Their product is (35), so (m=35).
Step 3
Exam Tip
योग (12) वाली अभाज्य जड़ें (5) और (7) हैं। उनका गुणनफल (35) है, इसलिए (m=35)।
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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 (x-2 -(2r+5)x+\(r^2+5r+6\)=0) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha-\beta\) का धनात्मक मान क्या है?
If \(\alpha,\beta\) are the roots of (x-2 -(2r+5)x+\(r^2+5r+6\)=0), what is the positive value of \(\alpha-\beta\)?
#quadratic-roots
#parametric-roots
#difference-of-roots
A (1)
B (2)
C (3)
D (5)
Explanation opens after your attempt
Step 1
Concept
In the given equation, the sum of roots is (2r+5) and the product is (r-2 +5r+6=(r+2)(r+3)). Hence the roots are (r+2) and (r+3), so the positive difference is (1).
Step 2
Why this answer is correct
The correct answer is A. (1). In the given equation, the sum of roots is (2r+5) and the product is (r-2 +5r+6=(r+2)(r+3)). Hence the roots are (r+2) and (r+3), so the positive difference is (1).
Step 3
Exam Tip
दिए गए समीकरण में जड़ों का योग (2r+5) और गुणनफल (r-2 +5r+6=(r+2)(r+3)) है। इसलिए जड़ें (r+2) और (r+3) हैं, अतः धनात्मक अंतर (1) है।
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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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यदि \(x^2+bx+c=0\) की जड़ें एक-दूसरे की विपरीत संख्याएँ हैं, तो कौन-सी शर्त अनिवार्य है?
If the roots of \(x^2+bx+c=0\) are opposites of each other, which condition is necessary?
#quadratic-roots
#opposite-roots
#sum-of-roots
A (b=0)
B (c=0)
C (b=c)
D \(b^2=4c\)
Explanation opens after your attempt
Step 1
Concept
Opposite roots have sum (0). Here the sum is (-b), so (b=0).
Step 2
Why this answer is correct
The correct answer is A. (b=0). Opposite roots have sum (0). Here the sum is (-b), so (b=0).
Step 3
Exam Tip
विपरीत जड़ों का योग (0) होता है। यहाँ योग (-b) है, इसलिए (b=0)।
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\(x^2-px+36=0\) की जड़ें धनात्मक पूर्णांक हैं और उनका अंतर (5) है, तो (p) का मान क्या है?
The roots of \(x^2-px+36=0\) are positive integers and their difference is (5). What is (p)?
#quadratic-roots
#integer-roots
#sum-of-roots
A (11)
B (12)
C (13)
D (15)
Explanation opens after your attempt
Step 1
Concept
The positive roots with product (36) and difference (5) are (4) and (9). Their sum is (13), so (p=13).
Step 2
Why this answer is correct
The correct answer is C. (13). The positive roots with product (36) and difference (5) are (4) and (9). Their sum is (13), so (p=13).
Step 3
Exam Tip
गुणनफल (36) और अंतर (5) वाली धनात्मक जड़ें (4) और (9) हैं। उनका योग (13) है, इसलिए (p=13)।
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यदि \(x^2-7x+k=0\) की जड़ें एक-दूसरे की व्युत्क्रम हैं, तो (k) का मान क्या होगा?
If the roots of \(x^2-7x+k=0\) are reciprocals of each other, what is the value of (k)?
#quadratic-roots
#reciprocal-roots
#product-of-roots
A (1)
B (-1)
C (7)
D (49)
Explanation opens after your attempt
Step 1
Concept
For reciprocal roots, the product is (1), and here the product is (k). Hence (k=1); in exams, check the product first.
Step 2
Why this answer is correct
The correct answer is A. (1). For reciprocal roots, the product is (1), and here the product is (k). Hence (k=1); in exams, check the product first.
Step 3
Exam Tip
व्युत्क्रम जड़ों के लिए गुणनफल (1) होता है और यहाँ गुणनफल (k) है। इसलिए (k=1); परीक्षा में पहले गुणनफल देखें।
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यदि \(4x^2-3x+k=0\) के मूलों का गुणनफल मूलों के योग के बराबर है तो (k) क्या होगा?
If the product of roots of \(4x^2-3x+k=0\) equals the sum of roots, what is (k)?
#roots
#sum_equals_product
#parameter
A (3)
B \(\frac{3}{4}\)
C (4)
D (1)
Explanation opens after your attempt
Step 1
Concept
The sum is \(-\frac{b}{a}=\frac{3}{4}\) and the product is \(\frac{k}{4}\). From \(\frac{k}{4}=\frac{3}{4}\), (k=3).
Step 2
Why this answer is correct
The correct answer is A. (3). The sum is \(-\frac{b}{a}=\frac{3}{4}\) and the product is \(\frac{k}{4}\). From \(\frac{k}{4}=\frac{3}{4}\), (k=3).
Step 3
Exam Tip
योग \(-\frac{b}{a}=\frac{3}{4}\) और गुणनफल \(\frac{k}{4}\) है। \(\frac{k}{4}=\frac{3}{4}\) से (k=3) है।
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यदि मूलों का योग (6) और उनके वर्गों का योग (52) है तो मूलों का गुणनफल क्या होगा?
If the sum of roots is (6) and the sum of their squares is (52), what is the product of roots?
#roots
#identity
#product
A (-8)
B (8)
C (16)
D (-16)
Explanation opens after your attempt
Step 1
Concept
(\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta). From \(52=36-2\alpha\beta\), we get \(\alpha\beta=-8\).
Step 2
Why this answer is correct
The correct answer is A. (-8). (\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta). From \(52=36-2\alpha\beta\), we get \(\alpha\beta=-8\).
Step 3
Exam Tip
(\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta) है। \(52=36-2\alpha\beta\) से \(\alpha\beta=-8\) मिलता है।
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यदि \(x^2-4x+3=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(3\alpha\) और \(3\beta\) को मूल मानकर समीकरण कौन सा होगा?
If \(\alpha\) and \(\beta\) are roots of \(x^2-4x+3=0\), which equation has \(3\alpha\) and \(3\beta\) as roots?
#roots
#transformed_roots
#equation
A \(x^2-12x+27=0\)
B \(x^2-4x+27=0\)
C \(x^2-12x+9=0\)
D \(x^2+12x+27=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-12x+27=0\)
Step 1
Concept
The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is \(x^2-12x+27=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-12x+27=0\). The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is \(x^2-12x+27=0\).
Step 3
Exam Tip
पुराने योग (4) और गुणनफल (3) हैं। नए योग (12) और गुणनफल (27) होंगे इसलिए \(x^2-12x+27=0\) है।
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यदि \(3x^2-2x+k=0\) के मूलों का गुणनफल मूलों के योग के बराबर है तो (k) क्या होगा?
If the product of roots of \(3x^2-2x+k=0\) equals the sum of roots, what is (k)?
#roots
#sum_equals_product
#parameter
A (2)
B \(\frac{2}{3}\)
C (3)
D (1)
Explanation opens after your attempt
Step 1
Concept
The sum is \(-\frac{b}{a}=\frac{2}{3}\) and the product is \(\frac{k}{3}\). From \(\frac{k}{3}=\frac{2}{3}\), (k=2).
Step 2
Why this answer is correct
The correct answer is A. (2). The sum is \(-\frac{b}{a}=\frac{2}{3}\) and the product is \(\frac{k}{3}\). From \(\frac{k}{3}=\frac{2}{3}\), (k=2).
Step 3
Exam Tip
योग \(-\frac{b}{a}=\frac{2}{3}\) और गुणनफल \(\frac{k}{3}\) है। \(\frac{k}{3}=\frac{2}{3}\) से (k=2) है।
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यदि मूलों का योग (4) और उनके वर्गों का योग (20) है तो मूलों का गुणनफल क्या होगा?
If the sum of roots is (4) and the sum of their squares is (20), what is the product of roots?
#roots
#identity
#product
A (-2)
B (2)
C (8)
D (-8)
Explanation opens after your attempt
Step 1
Concept
(\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta). From \(20=16-2\alpha\beta\), we get \(\alpha\beta=-2\).
Step 2
Why this answer is correct
The correct answer is A. (-2). (\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta). From \(20=16-2\alpha\beta\), we get \(\alpha\beta=-2\).
Step 3
Exam Tip
(\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta) है। \(20=16-2\alpha\beta\) से \(\alpha\beta=-2\) मिलता है।
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यदि \(x^2-3x+2=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(2\alpha\) और \(2\beta\) को मूल मानकर समीकरण कौन सा होगा?
If \(\alpha\) and \(\beta\) are roots of \(x^2-3x+2=0\), which equation has \(2\alpha\) and \(2\beta\) as roots?
#roots
#transformed_roots
#equation
A \(x^2-6x+8=0\)
B \(x^2-3x+8=0\)
C \(x^2-6x+4=0\)
D \(x^2+6x+8=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x+8=0\)
Step 1
Concept
The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is \(x^2-6x+8=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x+8=0\). The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is \(x^2-6x+8=0\).
Step 3
Exam Tip
पुराने योग (3) और गुणनफल (2) हैं। नए योग (6) और गुणनफल (8) होंगे इसलिए \(x^2-6x+8=0\) है।
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यदि किसी द्विघात समीकरण के मूलों का योग (0) है तो मूलों के बारे में सही कथन कौन सा हो सकता है?
If the sum of roots of a quadratic equation is (0), which statement about the roots can be correct?
#roots
#zero_sum
#reasoning
A मूल एक दूसरे के विपरीत हैं / The roots are opposites of each other
B दोनों मूल हमेशा (1) हैं / Both roots are always (1)
C दोनों मूल हमेशा धनात्मक हैं / Both roots are always positive
D मूलों का गुणनफल हमेशा (0) है / The product is always (0)
Explanation opens after your attempt
Correct Answer
A. मूल एक दूसरे के विपरीत हैं / The roots are opposites of each other
Step 1
Concept
If \(\alpha+\beta=0\), then \(\beta=-\alpha\). Therefore the roots can be opposites.
Step 2
Why this answer is correct
The correct answer is A. मूल एक दूसरे के विपरीत हैं / The roots are opposites of each other. If \(\alpha+\beta=0\), then \(\beta=-\alpha\). Therefore the roots can be opposites.
Step 3
Exam Tip
यदि \(\alpha+\beta=0\) है तो \(\beta=-\alpha\) होता है। इसलिए मूल विपरीत हो सकते हैं।
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यदि \(4\alpha\) और \(4\beta\) नए मूल हैं तथा \(\alpha+\beta=3\) है तो नए मूलों का योग क्या होगा?
If \(4\alpha\) and \(4\beta\) are new roots and \(\alpha+\beta=3\), what is the sum of the new roots?
#roots
#transformed_roots
#sum
A (12)
B (7)
C (3)
D \(\frac{3}{4}\)
Explanation opens after your attempt
Step 1
Concept
The sum of new roots is (4\alpha+4\beta=4\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 2
Why this answer is correct
The correct answer is A. (12). The sum of new roots is (4\alpha+4\beta=4\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 3
Exam Tip
नए मूलों का योग (4\alpha+4\beta=4\(\alpha+\beta\)=12) है। गुणक लगे मूलों में योग भी उसी गुणक से गुणा होता है।
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