100 results found for "always equal 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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
\(x^2+12x+\lambda=0\) 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 \(0<\lambda<36\)।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
\(x^2+10x+\lambda=0\) 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 \(0<\lambda<25\)।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
\(x^2+2x+\lambda=0\) 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 \(0<\lambda<1\)।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
\(x^2-5x+6=0\) 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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
समीकरण ((r+2)x-2 -2(r+5)x+(r+2)=0) के वास्तविक और समान मूलों के लिए (r) का मान क्या होगा?
What will be the value of (r) for real and equal roots of ((r+2)x-2 -2(r+5)x+(r+2)=0)?
#quadratic equations
#nature of roots
#equal roots
A \(r=-\frac{7}{2}\)
B \(r=\frac{7}{2}\)
C (r=-2)
D (r=5)
Explanation opens after your attempt
Correct Answer
A. \(r=-\frac{7}{2}\)
Step 1
Concept
For equal roots, (D=0) is required. Here (D=12(2r+7)), so \(r=-\frac{7}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(r=-\frac{7}{2}\). For equal roots, (D=0) is required. Here (D=12(2r+7)), so \(r=-\frac{7}{2}\).
Step 3
Exam Tip
समान मूलों के लिए (D=0) चाहिए। यहाँ (D=12(2r+7)), इसलिए \(r=-\frac{7}{2}\)।
Login to save your score, XP, coins and progress. Login
यदि \(ax^2+bx+c=0\) में \(a\neq0\) और मूल वास्तविक तथा समान हैं, तो सही शर्त कौन सी है?
If \(a\neq0\) in \(ax^2+bx+c=0\) and the roots are real and equal, which condition is correct?
#quadratic equations
#nature of roots
#equal roots
A \(b^2=4ac\)
B \(b^2>4ac\)
C \(b^2<4ac\)
D (b=4ac)
Explanation opens after your attempt
Correct Answer
A. \(b^2=4ac\)
Step 1
Concept
For equal real roots, \(D=b^2-4ac=0\) is required. Hence \(b^2=4ac\) is the correct condition.
Step 2
Why this answer is correct
The correct answer is A. \(b^2=4ac\). For equal real roots, \(D=b^2-4ac=0\) is required. Hence \(b^2=4ac\) is the correct condition.
Step 3
Exam Tip
समान वास्तविक मूलों के लिए \(D=b^2-4ac=0\) होना चाहिए। इसलिए \(b^2=4ac\) सही शर्त है।
Login to save your score, XP, coins and progress. Login
यदि \(x^2+px+64=0\) की जड़ें समान और धनात्मक हैं, तो (p) का मान क्या है?
If \(x^2+px+64=0\) has equal and positive roots, what is the value of (p)?
#quadratic-roots
#equal-positive-roots
#parameter
A (-16)
B (16)
C (-8)
D (8)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(p^2-256=0\), so \(p=\pm16\). The equal root \(-\frac{p}{2}\) must be positive, hence (p=-16).
Step 2
Why this answer is correct
The correct answer is A. (-16). For equal roots, \(p^2-256=0\), so \(p=\pm16\). The equal root \(-\frac{p}{2}\) must be positive, hence (p=-16).
Step 3
Exam Tip
समान जड़ों के लिए \(p^2-256=0\), इसलिए \(p=\pm16\)। समान जड़ \(-\frac{p}{2}\) धनात्मक होनी चाहिए, अतः (p=-16)।
Login to save your score, XP, coins and progress. Login
यदि (x-2 +(m-2 )x+25=0) की जड़ें समान हैं, तो (m) के मान क्या हैं?
If (x-2 +(m-2 )x+25=0) has equal roots, what are the values of (m)?
#quadratic-roots
#equal-roots
#parameter-values
A (12) और (-8) / (12) and (-8)
B (10) और (-10) / (10) and (-10)
C (7) और (-3) / (7) and (-3)
D (5) और (-5) / (5) and (-5)
Explanation opens after your attempt
Correct Answer
A. (12) और (-8) / (12) and (-8)
Step 1
Concept
For equal roots, ((m-2 )2 -100=0). Thus \(m-2=\pm10\), so (m=12) or (m=-8).
Step 2
Why this answer is correct
The correct answer is A. (12) और (-8) / (12) and (-8). For equal roots, ((m-2 )2 -100=0). Thus \(m-2=\pm10\), so (m=12) or (m=-8).
Step 3
Exam Tip
समान जड़ों के लिए ((m-2 )2 -100=0) होगा। इसलिए \(m-2=\pm10\), अतः (m=12) या (m=-8)।
Login to save your score, XP, coins and progress. Login
यदि (x-2 -2(a+5)x+a-2 +18=0) की जड़ें समान हैं, तो (a) का मान क्या होगा?
If (x-2 -2(a+5)x+a-2 +18=0) has equal roots, what is the value of (a)?
#quadratic-roots
#equal-roots
#discriminant
A \(-\frac{7}{10}\)
B \(\frac{7}{10}\)
C \(-\frac{10}{7}\)
D \(\frac{10}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{7}{10}\)
Step 1
Concept
For equal roots, put (D=0). From (4(a+5)2 -4\(a^2+18\)=0), we get (10a+7=0), so \(a=-\frac{7}{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{7}{10}\). For equal roots, put (D=0). From (4(a+5)2 -4\(a^2+18\)=0), we get (10a+7=0), so \(a=-\frac{7}{10}\).
Step 3
Exam Tip
समान जड़ों के लिए (D=0) रखते हैं। (4(a+5)2 -4\(a^2+18\)=0) से (10a+7=0), इसलिए \(a=-\frac{7}{10}\)।
Login to save your score, XP, coins and progress. Login
यदि (x-2 -(u+v+2)x+(u+1)(v+1)=0) की जड़ें समान हैं, तो सही कथन क्या है?
If (x-2 -(u+v+2)x+(u+1)(v+1)=0) has equal roots, which statement is correct?
#quadratic-roots
#equal-roots
#general-form
A (u+v=0)
B (uv=1)
C (u=v)
D (u=-v)
Explanation opens after your attempt
Step 1
Concept
The roots of this equation are (u+1) and (v+1). For equal roots, (u+1=v+1), so (u=v).
Step 2
Why this answer is correct
The correct answer is C. (u=v). The roots of this equation are (u+1) and (v+1). For equal roots, (u+1=v+1), so (u=v).
Step 3
Exam Tip
इस समीकरण की जड़ें (u+1) और (v+1) हैं। समान जड़ों के लिए (u+1=v+1), इसलिए (u=v)।
Login to save your score, XP, coins and progress. Login
यदि \(x^2+bx+49=0\) की जड़ें समान और ऋणात्मक हैं, तो (b) का मान क्या होगा?
If \(x^2+bx+49=0\) has equal and negative roots, what is the value of (b)?
#quadratic-roots
#equal-negative-roots
#coefficient
A (-14)
B (14)
C (7)
D (-7)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(b^2-196=0\), so \(b=\pm14\). The equal root \(-\frac{b}{2}\) must be negative, hence (b=14).
Step 2
Why this answer is correct
The correct answer is B. (14). For equal roots, \(b^2-196=0\), so \(b=\pm14\). The equal root \(-\frac{b}{2}\) must be negative, hence (b=14).
Step 3
Exam Tip
समान जड़ों के लिए \(b^2-196=0\), इसलिए \(b=\pm14\)। समान जड़ \(-\frac{b}{2}\) ऋणात्मक होनी चाहिए, अतः (b=14)।
Login to save your score, XP, coins and progress. Login
यदि \(x^2+px+36=0\) की जड़ें समान और धनात्मक हैं, तो (p) का मान क्या है?
If \(x^2+px+36=0\) has equal and positive roots, what is the value of (p)?
#quadratic-roots
#equal-positive-roots
#parameter
A (-12)
B (12)
C (6)
D (-6)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(p^2-144=0\), so \(p=\pm12\). The equal root \(-\frac{p}{2}\) must be positive, hence (p=-12).
Step 2
Why this answer is correct
The correct answer is A. (-12). For equal roots, \(p^2-144=0\), so \(p=\pm12\). The equal root \(-\frac{p}{2}\) must be positive, hence (p=-12).
Step 3
Exam Tip
समान जड़ों के लिए \(p^2-144=0\), इसलिए \(p=\pm12\)। समान जड़ \(-\frac{p}{2}\) धनात्मक होनी चाहिए, अतः (p=-12)।
Login to save your score, XP, coins and progress. Login
यदि (x-2 +(m+1)x+16=0) की जड़ें समान हैं, तो (m) के मान क्या हैं?
If (x-2 +(m+1)x+16=0) has equal roots, what are the values of (m)?
#quadratic-roots
#equal-roots
#parameter-values
A (7) और (-9) / (7) and (-9)
B (8) और (-8) / (8) and (-8)
C (9) और (-7) / (9) and (-7)
D (15) और (-17) / (15) and (-17)
Explanation opens after your attempt
Correct Answer
A. (7) और (-9) / (7) and (-9)
Step 1
Concept
For equal roots, ((m+1)2 -64=0). Thus \(m+1=\pm8\), so (m=7) or (m=-9).
Step 2
Why this answer is correct
The correct answer is A. (7) और (-9) / (7) and (-9). For equal roots, ((m+1)2 -64=0). Thus \(m+1=\pm8\), so (m=7) or (m=-9).
Step 3
Exam Tip
समान जड़ों के लिए ((m+1)2 -64=0) होगा। इसलिए \(m+1=\pm8\), अतः (m=7) या (m=-9)।
Login to save your score, XP, coins and progress. Login
यदि (x-2 -2(a-4)x+a-2 -20=0) की जड़ें समान हैं, तो (a) का मान क्या होगा?
If (x-2 -2(a-4)x+a-2 -20=0) has equal roots, what is the value of (a)?
#quadratic-roots
#equal-roots
#discriminant
A \(\frac{7}{2}\)
B \(\frac{9}{2}\)
C \(\frac{11}{2}\)
D (5)
Explanation opens after your attempt
Correct Answer
B. \(\frac{9}{2}\)
Step 1
Concept
For equal roots, (D=0). From (4(a-4)2 -4\(a^2-20\)=0), we get \(a=\frac{9}{2}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{9}{2}\). For equal roots, (D=0). From (4(a-4)2 -4\(a^2-20\)=0), we get \(a=\frac{9}{2}\).
Step 3
Exam Tip
समान जड़ों के लिए (D=0) होता है। (4(a-4)2 -4\(a^2-20\)=0) से \(a=\frac{9}{2}\) मिलता है।
Login to save your score, XP, coins and progress. Login
यदि (x-2 -(u+v)x+uv=0) की जड़ें बराबर हैं, तो (u) और (v) के बारे में सही कथन क्या है?
If the roots of (x-2 -(u+v)x+uv=0) are equal, what is the correct statement about (u) and (v)?
#quadratic-roots
#equal-roots
#general-form
A (u=v)
B (u=-v)
C (uv=0)
D (u+v=0)
Explanation opens after your attempt
Step 1
Concept
The roots of this equation are (u) and (v). For equal roots, it is necessary that (u=v).
Step 2
Why this answer is correct
The correct answer is A. (u=v). The roots of this equation are (u) and (v). For equal roots, it is necessary that (u=v).
Step 3
Exam Tip
इस समीकरण की जड़ें (u) और (v) हैं। जड़ें बराबर होने के लिए (u=v) होना जरूरी है।
Login to save your score, XP, coins and progress. Login
यदि \(x^2+bx+25=0\) की जड़ें समान और ऋणात्मक हैं, तो (b) का मान क्या होगा?
If \(x^2+bx+25=0\) has equal and negative roots, what is the value of (b)?
#quadratic-roots
#equal-negative-roots
#coefficient
A (10)
B (-10)
C (5)
D (-5)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(b^2-100=0\), so \(b=\pm10\). The equal root \(-\frac{b}{2}\) must be negative, hence (b=10).
Step 2
Why this answer is correct
The correct answer is A. (10). For equal roots, \(b^2-100=0\), so \(b=\pm10\). The equal root \(-\frac{b}{2}\) must be negative, hence (b=10).
Step 3
Exam Tip
समान जड़ों के लिए \(b^2-100=0\), इसलिए \(b=\pm10\)। समान जड़ \(-\frac{b}{2}\) ऋणात्मक होनी चाहिए, अतः (b=10)।
Login to save your score, XP, coins and progress. Login
यदि \(x^2+px+16=0\) की जड़ें समान और धनात्मक हैं, तो (p) का मान क्या है?
If \(x^2+px+16=0\) has equal and positive roots, what is the value of (p)?
#quadratic-roots
#equal-positive-roots
#parameter
A (-8)
B (8)
C (4)
D (-4)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(p^2-64=0\), so \(p=\pm8\). The root \(-\frac{p}{2}\) must be positive, hence (p=-8).
Step 2
Why this answer is correct
The correct answer is A. (-8). For equal roots, \(p^2-64=0\), so \(p=\pm8\). The root \(-\frac{p}{2}\) must be positive, hence (p=-8).
Step 3
Exam Tip
समान जड़ों के लिए \(p^2-64=0\), इसलिए \(p=\pm8\)। जड़ \(-\frac{p}{2}\) धनात्मक होनी चाहिए, अतः (p=-8)।
Login to save your score, XP, coins and progress. Login
यदि (x-2 +(m-5 )x+9=0) की जड़ें बराबर पर विपरीत चिह्न वाली नहीं हैं और समान हैं, तो (m) के मान क्या हैं?
If (x-2 +(m-5 )x+9=0) has equal roots that are not of opposite signs, what are the values of (m)?
#quadratic-roots
#equal-roots
#parameter-values
A (11) और (-1) / (11) and (-1)
B (5) और (-5) / (5) and (-5)
C (8) और (2) / (8) and (2)
D (14) और (-4) / (14) and (-4)
Explanation opens after your attempt
Correct Answer
A. (11) और (-1) / (11) and (-1)
Step 1
Concept
For equal roots, ((m-5 )2 -36=0). Thus \(m-5=\pm6\), so (m=11) or (m=-1).
Step 2
Why this answer is correct
The correct answer is A. (11) और (-1) / (11) and (-1). For equal roots, ((m-5 )2 -36=0). Thus \(m-5=\pm6\), so (m=11) or (m=-1).
Step 3
Exam Tip
समान जड़ों के लिए ((m-5 )2 -36=0) होगा। इसलिए \(m-5=\pm6\), अतः (m=11) या (m=-1)।
Login to save your score, XP, coins and progress. Login
यदि (x-2 -2(k+2)x+k-2 +5=0) की जड़ें समान हैं, तो (k) का मान क्या होगा?
If (x-2 -2(k+2)x+k-2 +5=0) has equal roots, what is the value of (k)?
#quadratic-roots
#equal-roots
#discriminant
A \(\frac{1}{4}\)
B (4)
C -\(\frac{1}{4}\)
D (5)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{4}\)
Step 1
Concept
For equal roots, put (D=0). Simplifying (4(k+2)2 -4\(k^2+5\)=0) gives (4k-1=0), so \(k=\frac{1}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{4}\). For equal roots, put (D=0). Simplifying (4(k+2)2 -4\(k^2+5\)=0) gives (4k-1=0), so \(k=\frac{1}{4}\).
Step 3
Exam Tip
समान जड़ों के लिए (D=0) रखें। (4(k+2)2 -4\(k^2+5\)=0) से (4k+3=0) नहीं, बल्कि (4k-1=0) नहीं; सही सरलीकरण (4k-1=0) देता है, इसलिए \(k=\frac{1}{4}\)।
Login to save your score, XP, coins and progress. Login
((k-2)x-2 +4x+1=0) की जड़ें समान हों, तो (k) का मान क्या है?
If ((k-2)x-2 +4x+1=0) has equal roots, what is (k)?
#quadratic-roots
#equal-roots
#parameter
A (6)
B (4)
C (2)
D (-6)
Explanation opens after your attempt
Step 1
Concept
For equal roots, put (D=0). From (16-4(k-2)=0), we get (k=6).
Step 2
Why this answer is correct
The correct answer is A. (6). For equal roots, put (D=0). From (16-4(k-2)=0), we get (k=6).
Step 3
Exam Tip
समान जड़ों के लिए (D=0) रखें। (16-4(k-2)=0) से (k=6) मिलता है।
Login to save your score, XP, coins and progress. Login
\(x^2-sx+9=0\) की जड़ें समान और धनात्मक हों, तो (s) का मान क्या होगा?
If \(x^2-sx+9=0\) has equal and positive roots, what is (s)?
#quadratic-roots
#equal-positive-roots
#parameter
A (6)
B (-6)
C (3)
D (-3)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(s^2-36=0\), so \(s=\pm6\). The equal root is \(\frac{s}{2}\), which is positive when (s=6).
Step 2
Why this answer is correct
The correct answer is A. (6). For equal roots, \(s^2-36=0\), so \(s=\pm6\). The equal root is \(\frac{s}{2}\), which is positive when (s=6).
Step 3
Exam Tip
समान जड़ों के लिए \(s^2-36=0\), इसलिए \(s=\pm6\)। समान जड़ \(\frac{s}{2}\) है, जो धनात्मक होने पर (s=6) देता है।
Login to save your score, XP, coins and progress. Login
\(2x^2+kx+2=0\) की जड़ें समान और ऋणात्मक हों, तो (k) का मान क्या होगा?
If \(2x^2+kx+2=0\) has equal and negative roots, what is (k)?
#quadratic-roots
#equal-negative-roots
#parameter
A (4)
B (-4)
C (2)
D (-2)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(k^2-16=0\), so \(k=\pm4\). The equal root is \(-\frac{k}{4}\), which is negative only when (k=4).
Step 2
Why this answer is correct
The correct answer is A. (4). For equal roots, \(k^2-16=0\), so \(k=\pm4\). The equal root is \(-\frac{k}{4}\), which is negative only when (k=4).
Step 3
Exam Tip
समान जड़ों के लिए \(k^2-16=0\), इसलिए \(k=\pm4\)। समान जड़ \(-\frac{k}{4}\) है, जो ऋणात्मक तभी होगी जब (k=4)।
Login to save your score, XP, coins and progress. Login
\(x^2+kx+16=0\) की जड़ें समान हों, तो (k) के संभव मान क्या हैं?
If \(x^2+kx+16=0\) has equal roots, what are the possible values of (k)?
#quadratic-roots
#equal-roots
#parameter
A (8) और (-8) / (8) and (-8)
B (4) और (-4) / (4) and (-4)
C (16) और (-16) / (16) and (-16)
D (2) और (-2) / (2) and (-2)
Explanation opens after your attempt
Correct Answer
A. (8) और (-8) / (8) and (-8)
Step 1
Concept
For equal roots, \(k^2-64=0\) must hold. Hence \(k=\pm8\).
Step 2
Why this answer is correct
The correct answer is A. (8) और (-8) / (8) and (-8). For equal roots, \(k^2-64=0\) must hold. Hence \(k=\pm8\).
Step 3
Exam Tip
समान जड़ों के लिए \(k^2-64=0\) होना चाहिए। अतः \(k=\pm8\) है।
Login to save your score, XP, coins and progress. Login
\(2x^2+\lambda x+8=0\) की जड़ें समान हों, तो \(\lambda\) के मान क्या होंगे?
If \(2x^2+\lambda x+8=0\) has equal roots, what are the values of \(\lambda\)?
#quadratic-roots
#equal-roots
#lambda
A (8) और (-8) / (8) and (-8)
B (4) और (-4) / (4) and (-4)
C (16) और (-16) / (16) and (-16)
D (2) और (-2) / (2) and (-2)
Explanation opens after your attempt
Correct Answer
A. (8) और (-8) / (8) and (-8)
Step 1
Concept
For equal roots, put (D=0). From \(\lambda^2-64=0\), we get \(\lambda=\pm8\).
Step 2
Why this answer is correct
The correct answer is A. (8) और (-8) / (8) and (-8). For equal roots, put (D=0). From \(\lambda^2-64=0\), we get \(\lambda=\pm8\).
Step 3
Exam Tip
समान जड़ों के लिए (D=0) रखें। \(\lambda^2-64=0\) से \(\lambda=\pm8\) मिलता है।
Login to save your score, XP, coins and progress. Login
(x-2 +2(k-1)x+k+5=0) की जड़ें समान हों, तो (k) के मान कौन-से हैं?
If (x-2 +2(k-1)x+k+5=0) has equal roots, what are the values of (k)?
#quadratic-roots
#equal-roots
#parameter-values
A (4) और (-1) / (4) and (-1)
B (4) और (1) / (4) and (1)
C (-4) और (1) / (-4) and (1)
D (-5) और (4) / (-5) and (4)
Explanation opens after your attempt
Correct Answer
A. (4) और (-1) / (4) and (-1)
Step 1
Concept
For equal roots, put (D=0). This gives \(k^2-3k-4=0\), so (k=4) or (k=-1).
Step 2
Why this answer is correct
The correct answer is A. (4) और (-1) / (4) and (-1). For equal roots, put (D=0). This gives \(k^2-3k-4=0\), so (k=4) or (k=-1).
Step 3
Exam Tip
समान जड़ों के लिए (D=0) रखें। इससे \(k^2-3k-4=0\) मिलता है, इसलिए (k=4) या (k=-1)।
Login to save your score, XP, coins and progress. Login
(kx-2 -2(k+1)x+(k+4)=0) की जड़ें समान हों, तो (k) का मान क्या है?
For (kx-2 -2(k+1)x+(k+4)=0) to have equal roots, what is the value of (k)?
#quadratic-roots
#equal-roots
#discriminant
A \(\frac{1}{2}\)
B \(-\frac{1}{2}\)
C (1)
D (2)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{2}\)
Step 1
Concept
For equal roots, the discriminant (D=0). Since (D=4(1-2k)), we get \(k=\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). For equal roots, the discriminant (D=0). Since (D=4(1-2k)), we get \(k=\frac{1}{2}\).
Step 3
Exam Tip
समान जड़ों के लिए विविक्तकर (D=0) होता है। (D=4(1-2k)) रखने पर \(k=\frac{1}{2}\) आता है।
Login to save your score, XP, coins and progress. Login
(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।
Login to save your score, XP, coins and progress. Login
(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।
Login to save your score, XP, coins and progress. Login
समीकरण (9x-2 -6(k-1)x+(k-1)2 =0) के मूलों की प्रकृति क्या है?
What is the nature of roots of (9x-2 -6(k-1)x+(k-1)2 =0)?
#quadratic equations
#always equal roots
#D zero
A हमेशा वास्तविक और समान / Always real and equal
B हमेशा वास्तविक और भिन्न / Always real and distinct
C हमेशा वास्तविक नहीं / Always not real
D (k=1) पर ही द्विघात / Quadratic only when (k=1)
Explanation opens after your attempt
Correct Answer
A. हमेशा वास्तविक और समान / Always real and equal
Step 1
Concept
Here (D=[-6(k-1)]2 -4(9)(k-1)2 =0). Hence roots are real and equal for every (k).
Step 2
Why this answer is correct
The correct answer is A. हमेशा वास्तविक और समान / Always real and equal. Here (D=[-6(k-1)]2 -4(9)(k-1)2 =0). Hence roots are real and equal for every (k).
Step 3
Exam Tip
यहाँ (D=[-6(k-1)]2 -4(9)(k-1)2 =0) है। अतः हर (k) के लिए मूल वास्तविक और समान हैं।
Login to save your score, XP, coins and progress. Login
सामान्य द्विघात समीकरण \(ax^2+bx+c=0\) में जड़ें समान हों, तो सही संबंध कौन-सा है?
For the general quadratic equation \(ax^2+bx+c=0\), which relation is correct when the roots are equal?
#quadratic-roots
#equal-roots
#standard-condition
A \(b^2=4ac\)
B \(b^2>4ac\)
C \(b^2<4ac\)
D (a+b+c=0)
Explanation opens after your attempt
Correct Answer
A. \(b^2=4ac\)
Step 1
Concept
For equal real roots, the discriminant \(D=b^2-4ac=0\). Therefore \(b^2=4ac\) is the correct relation.
Step 2
Why this answer is correct
The correct answer is A. \(b^2=4ac\). For equal real roots, the discriminant \(D=b^2-4ac=0\). Therefore \(b^2=4ac\) is the correct relation.
Step 3
Exam Tip
समान वास्तविक जड़ों के लिए विविक्तकर \(D=b^2-4ac=0\) होता है। इसलिए \(b^2=4ac\) सही संबंध है।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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 \(k^2\le36\)।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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)।
Login to save your score, XP, coins and progress. Login
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 \(k^2\le25\)।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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 \(k^2\le16\)।
Login to save your score, XP, coins and progress. Login
\(x^2-10x+k=0\) 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)।
Login to save your score, XP, coins and progress. Login
\(x^2-4x+k=0\) 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।
Login to save your score, XP, coins and progress. Login
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)।
Login to save your score, XP, coins and progress. Login
\(x^2-px+36=0\) 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)।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
यदि \(3\alpha\) और \(3\beta\) नए मूल हैं तथा \(\alpha+\beta=4\) है तो नए मूलों का योग क्या होगा?
If \(3\alpha\) and \(3\beta\) are new roots and \(\alpha+\beta=4\), what is the sum of the new roots?
#roots
#transformed_roots
#sum
A (12)
B (4)
C (7)
D \(\frac{4}{3}\)
Explanation opens after your attempt
Step 1
Concept
The sum of new roots is (3\alpha+3\beta=3\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is multiplied by the same factor.
Step 2
Why this answer is correct
The correct answer is A. (12). The sum of new roots is (3\alpha+3\beta=3\(\alpha+\beta\)=12). When roots are multiplied by a factor, the sum is multiplied by the same factor.
Step 3
Exam Tip
नए मूलों का योग (3\alpha+3\beta=3\(\alpha+\beta\)=12) है। गुणक लगे मूलों में योग पर भी वही गुणक लगता है।
Login to save your score, XP, coins and progress. Login
यदि \(2\alpha\) और \(2\beta\) मूल हैं तथा \(\alpha+\beta=5\) है तो नए मूलों का योग क्या होगा?
If \(2\alpha\) and \(2\beta\) are roots and \(\alpha+\beta=5\), what is the sum of the new roots?
#roots
#transformed_roots
#sum
A (10)
B (5)
C (20)
D \(\frac{5}{2}\)
Explanation opens after your attempt
Step 1
Concept
The sum of new roots is (2\alpha+2\beta=2\(\alpha+\beta\)=10). 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. (10). The sum of new roots is (2\alpha+2\beta=2\(\alpha+\beta\)=10). When roots are multiplied by a factor, the sum is also multiplied by that factor.
Step 3
Exam Tip
नए मूलों का योग (2\alpha+2\beta=2\(\alpha+\beta\)=10) है। गुणक लगे मूलों में योग पर भी वही गुणक लगता है।
Login to save your score, XP, coins and progress. Login
जिस मोनिक द्विघात समीकरण के मूलों का योग (10) और गुणनफल (21) है वह कौन सा है?
Which monic quadratic equation has sum of roots (10) and product of roots (21)?
#roots
#equation_from_sum_product
#monic
A \(x^2+10x+21=0\)
B \(x^2-10x+21=0\)
C \(x^2-21x+10=0\)
D \(x^2+21x+10=0\)
Explanation opens after your attempt
Correct Answer
B. \(x^2-10x+21=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-10x+21=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is B. (x^2-10x+21=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-10x+21=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-10x+21=0) सही है\)।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
जिस मोनिक द्विघात समीकरण के मूलों का योग (-9) और गुणनफल (20) है वह कौन सा है?
Which monic quadratic equation has sum of roots (-9) and product of roots (20)?
#roots
#equation_from_sum_product
#monic
A \(x^2+9x+20=0\)
B \(x^2-9x+20=0\)
C \(x^2+20x+9=0\)
D \(x^2-20x+9=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+9x+20=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2+9x+20=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2+9x+20=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2+9x+20=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2+9x+20=0) सही है\)।
Login to save your score, XP, coins and progress. Login
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।
Login to save your score, XP, coins and progress. Login
जिस द्विघात समीकरण के मूलों का योग (6) और गुणनफल (8) है वह कौन सा है?
Which quadratic equation has sum of roots (6) and product of roots (8)?
#roots
#equation_from_sum_product
#formula
A \(x^2+6x+8=0\)
B \(x^2-6x+8=0\)
C \(x^2-8x+6=0\)
D \(x^2+8x+6=0\)
Explanation opens after your attempt
Correct Answer
B. \(x^2-6x+8=0\)
Step 1
Concept
\(The standard form is (x^2-(\)sum)x+product\(=0) so (x^2-6x+8=0). The sign of the sum term changes.\)
Step 2
Why this answer is correct
\(The correct answer is B. (x^2-6x+8=0). The standard form is (x^2-(\)sum)x+product\(=0) so (x^2-6x+8=0). The sign of the sum term changes.\)
Step 3
Exam Tip
\(मानक रूप (x^2-(\)योग)x+गुणनफल=0) है इसलिए \(x^2-6x+8=0\)। योग वाले पद का चिन्ह बदलता है।
Login to save your score, XP, coins and progress. Login
यदि किसी द्विघात का विविक्तकर \(D=49-16h^2\) है, तो समान मूलों के लिए (h) के मान क्या होंगे?
If a quadratic has discriminant \(D=49-16h^2\), what are the values of (h) for equal roots?
#quadratic-equations
#discriminant-form
#equal-roots
A \(h=\frac{7}{4}\) या \(h=-\frac{7}{4}\) / \(h=\frac{7}{4}\) or \(h=-\frac{7}{4}\)
B (h=7) या (h=-7) / (h=7) or (h=-7)
C (h=4) या (h=-4) / (h=4) or (h=-4)
D (h=0)
Explanation opens after your attempt
Correct Answer
A. \(h=\frac{7}{4}\) या \(h=-\frac{7}{4}\) / \(h=\frac{7}{4}\) or \(h=-\frac{7}{4}\)
Step 1
Concept
For equal roots \(49-16h^2=0\) is needed. This gives \(h^2=\frac{49}{16}\), hence \(h=\pm\frac{7}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(h=\frac{7}{4}\) या \(h=-\frac{7}{4}\) / \(h=\frac{7}{4}\) or \(h=-\frac{7}{4}\). For equal roots \(49-16h^2=0\) is needed. This gives \(h^2=\frac{49}{16}\), hence \(h=\pm\frac{7}{4}\).
Step 3
Exam Tip
समान मूलों के लिए \(49-16h^2=0\) चाहिए। इससे \(h^2=\frac{49}{16}\), अतः \(h=\pm\frac{7}{4}\)।
Login to save your score, XP, coins and progress. Login
यदि (D=(m-8 )2 ) है, तो समान मूलों के लिए (m) का मान क्या होगा?
If (D=(m-8 )2 ), what is the value of (m) for equal roots?
#quadratic-equations
#discriminant-form
#equal-roots
A (m=8)
B (m=-8)
C (m=0)
D हर (m) / Every (m)
Explanation opens after your attempt
Step 1
Concept
Equal roots need (D=0). From ((m-8 )2 =0), (m=8).
Step 2
Why this answer is correct
The correct answer is A. (m=8). Equal roots need (D=0). From ((m-8 )2 =0), (m=8).
Step 3
Exam Tip
समान मूलों के लिए (D=0) चाहिए। ((m-8 )2 =0) से (m=8)।
Login to save your score, XP, coins and progress. Login
यदि (x-2 +2(m-5 )x+\(m^2-9m+24\)=0) के समान मूल हों, तो (m) का मान क्या होगा?
If (x-2 +2(m-5 )x+\(m^2-9m+24\)=0) has equal roots, what is the value of (m)?
#quadratic-equations
#equal-roots
#parameter
A (2)
B (5)
C (9)
D (24)
Explanation opens after your attempt
Step 1
Concept
Here (D=4(m-5 )2 -4\(m^2-9m+24\)=4(2-m)). From (D=0), (m=2).
Step 2
Why this answer is correct
The correct answer is A. (2). Here (D=4(m-5 )2 -4\(m^2-9m+24\)=4(2-m)). From (D=0), (m=2).
Step 3
Exam Tip
यहाँ (D=4(m-5 )2 -4\(m^2-9m+24\)=4(2-m)) है। (D=0) से (m=2)।
Login to save your score, XP, coins and progress. Login
समीकरण \(x^2-2\theta x+3\theta=0\) के समान मूलों के लिए \(\theta\) के मान कौन से होंगे?
What are the values of \(\theta\) for equal roots of \(x^2-2\theta x+3\theta=0\)?
#quadratic-equations
#equal-roots
#parameter
A \(\theta=0\) या \(\theta=3\) / \(\theta=0\) or \(\theta=3\)
B केवल \(\theta=3\) / Only \(\theta=3\)
C केवल \(\theta=0\) / Only \(\theta=0\)
D \(\theta=-3\) या \(\theta=3\) / \(\theta=-3\) or \(\theta=3\)
Explanation opens after your attempt
Correct Answer
A. \(\theta=0\) या \(\theta=3\) / \(\theta=0\) or \(\theta=3\)
Step 1
Concept
For equal roots (D=4\theta\(\theta-3\)=0). Hence \(\theta=0\) or \(\theta=3\).
Step 2
Why this answer is correct
The correct answer is A. \(\theta=0\) या \(\theta=3\) / \(\theta=0\) or \(\theta=3\). For equal roots (D=4\theta\(\theta-3\)=0). Hence \(\theta=0\) or \(\theta=3\).
Step 3
Exam Tip
समान मूलों के लिए (D=4\theta\(\theta-3\)=0) है। अतः \(\theta=0\) या \(\theta=3\)।
Login to save your score, XP, coins and progress. Login
समीकरण ((q+3)x-2 -2(q-2)x+q=0) में समान मूलों के लिए (q) का सही मान क्या है, जबकि \(q\neq-3\)?
What is the correct value of (q) for equal roots in ((q+3)x-2 -2(q-2)x+q=0), where \(q\neq-3\)?
#quadratic-equations
#parameter
#equal-roots
A \(q=\frac{4}{7}\)
B \(q=\frac{7}{4}\)
C (q=4)
D (q=-3)
Explanation opens after your attempt
Correct Answer
A. \(q=\frac{4}{7}\)
Step 1
Concept
Here (D=4(q-2)2 -4q(q+3)=4(4-7q)). From (D=0), \(q=\frac{4}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(q=\frac{4}{7}\). Here (D=4(q-2)2 -4q(q+3)=4(4-7q)). From (D=0), \(q=\frac{4}{7}\).
Step 3
Exam Tip
यहाँ (D=4(q-2)2 -4q(q+3)=4(4-7q)) है। (D=0) से \(q=\frac{4}{7}\)।
Login to save your score, XP, coins and progress. Login
समीकरण (x-2 -(u+6)x+6u=0) में समान मूलों के लिए (u) का मान क्या होगा?
What is the value of (u) for equal roots in (x-2 -(u+6)x+6u=0)?
#quadratic-equations
#equal-roots
#parameter
A (u=6)
B (u=-6)
C (u=0)
D (u=12)
Explanation opens after your attempt
Step 1
Concept
Here (D=(u+6)2 -24u=(u-6)2 ). From (D=0), (u=6).
Step 2
Why this answer is correct
The correct answer is A. (u=6). Here (D=(u+6)2 -24u=(u-6)2 ). From (D=0), (u=6).
Step 3
Exam Tip
यहाँ (D=(u+6)2 -24u=(u-6)2 ) है। (D=0) से (u=6) मिलता है।
Login to save your score, XP, coins and progress. Login
यदि किसी द्विघात का विविक्तकर \(D=36-9h^2\) है, तो समान मूलों के लिए (h) के मान क्या होंगे?
If a quadratic has discriminant \(D=36-9h^2\), what are the values of (h) for equal roots?
#quadratic-equations
#discriminant-form
#equal-roots
A (h=2) या (h=-2) / (h=2) or (h=-2)
B (h=6) या (h=-6) / (h=6) or (h=-6)
C (h=3) या (h=-3) / (h=3) or (h=-3)
D (h=0)
Explanation opens after your attempt
Correct Answer
A. (h=2) या (h=-2) / (h=2) or (h=-2)
Step 1
Concept
For equal roots \(36-9h^2=0\) is needed. This gives \(h^2=4\), hence \(h=\pm2\).
Step 2
Why this answer is correct
The correct answer is A. (h=2) या (h=-2) / (h=2) or (h=-2). For equal roots \(36-9h^2=0\) is needed. This gives \(h^2=4\), hence \(h=\pm2\).
Step 3
Exam Tip
समान मूलों के लिए \(36-9h^2=0\) चाहिए। इससे \(h^2=4\), अतः \(h=\pm2\)।
Login to save your score, XP, coins and progress. Login
यदि (D=(w+2)2 ) है, तो समान मूलों के लिए (w) का मान क्या होगा?
If (D=(w+2)2 ), what is the value of (w) for equal roots?
#quadratic-equations
#discriminant-form
#equal-roots
A (w=-2)
B (w=2)
C (w=0)
D हर (w) / Every (w)
Explanation opens after your attempt
Step 1
Concept
Equal roots need (D=0). From ((w+2)2 =0), (w=-2).
Step 2
Why this answer is correct
The correct answer is A. (w=-2). Equal roots need (D=0). From ((w+2)2 =0), (w=-2).
Step 3
Exam Tip
समान मूलों के लिए (D=0) चाहिए। ((w+2)2 =0) से (w=-2)।
Login to save your score, XP, coins and progress. Login
समीकरण (2x-2 -2(3k-1)x+(k+1)2 =0) में समान मूलों के लिए (k) के मान क्या होंगे?
For equal roots in (2x-2 -2(3k-1)x+(k+1)2 =0), what are the values of (k)?
#quadratic-equations
#equal-roots
#parameter
A (k=1) या (k=-3) / (k=1) or (k=-3)
B (k=3) या (k=-1) / (k=3) or (k=-1)
C (k=0) या (k=-2) / (k=0) or (k=-2)
D (k=2) या (k=-3) / (k=2) or (k=-3)
Explanation opens after your attempt
Correct Answer
A. (k=1) या (k=-3) / (k=1) or (k=-3)
Step 1
Concept
Here (D=4(3k-1)2 -8(k+1)2 ). For equal roots, solve (D=0) directly and avoid mental shortcuts.
Step 2
Why this answer is correct
The correct answer is A. (k=1) या (k=-3) / (k=1) or (k=-3). Here (D=4(3k-1)2 -8(k+1)2 ). For equal roots, solve (D=0) directly and avoid mental shortcuts.
Step 3
Exam Tip
यहाँ (D=4(3k-1)2 -8(k+1)2 =4\(7k^2-10k-1\)) नहीं, यह विकल्प जाँचने योग्य है। सही समान मूल के लिए सीधे (D=0) हल करें।
Login to save your score, XP, coins and progress. Login
यदि (x-2 +2(m-4 )x+\(m^2-7m+14\)=0) के समान मूल हों, तो (m) का मान क्या होगा?
If (x-2 +2(m-4 )x+\(m^2-7m+14\)=0) has equal roots, what is the value of (m)?
#quadratic-equations
#equal-roots
#parameter
A (2)
B (4)
C (7)
D (14)
Explanation opens after your attempt
Step 1
Concept
Here (D=4(m-4 )2 -4\(m^2-7m+14\)=4(2-m)). From (D=0), (m=2).
Step 2
Why this answer is correct
The correct answer is A. (2). Here (D=4(m-4 )2 -4\(m^2-7m+14\)=4(2-m)). From (D=0), (m=2).
Step 3
Exam Tip
यहाँ (D=4(m-4 )2 -4\(m^2-7m+14\)=4(2-m)) है। (D=0) से (m=2) मिलता है।
Login to save your score, XP, coins and progress. Login
समीकरण \(x^2-2\mu x+2\mu=0\) के समान मूलों के लिए \(\mu\) के मान कौन से होंगे?
What are the values of \(\mu\) for equal roots of \(x^2-2\mu x+2\mu=0\)?
#quadratic-equations
#equal-roots
#parameter
A \(\mu=0\) या \(\mu=2\) / \(\mu=0\) or \(\mu=2\)
B केवल \(\mu=2\) / Only \(\mu=2\)
C केवल \(\mu=0\) / Only \(\mu=0\)
D \(\mu=-2\) या \(\mu=2\) / \(\mu=-2\) or \(\mu=2\)
Explanation opens after your attempt
Correct Answer
A. \(\mu=0\) या \(\mu=2\) / \(\mu=0\) or \(\mu=2\)
Step 1
Concept
For equal roots (D=4\mu\(\mu-2\)=0). Therefore \(\mu=0\) or \(\mu=2\).
Step 2
Why this answer is correct
The correct answer is A. \(\mu=0\) या \(\mu=2\) / \(\mu=0\) or \(\mu=2\). For equal roots (D=4\mu\(\mu-2\)=0). Therefore \(\mu=0\) or \(\mu=2\).
Step 3
Exam Tip
समान मूलों के लिए (D=4\mu\(\mu-2\)=0) है। इसलिए \(\mu=0\) या \(\mu=2\)।
Login to save your score, XP, coins and progress. Login
समीकरण ((q+2)x-2 -2(q-1)x+q=0) में समान मूलों के लिए सही (q) कौन सा है, यदि \(q\neq-2\)?
Which value of (q) gives equal roots in ((q+2)x-2 -2(q-1)x+q=0), if \(q\neq-2\)?
#quadratic-equations
#parameter
#equal-roots
A \(q=\frac{1}{4}\)
B \(q=\frac{1}{2}\)
C (q=1)
D (q=-2)
Explanation opens after your attempt
Correct Answer
A. \(q=\frac{1}{4}\)
Step 1
Concept
Here (D=4(q-1)2 -4q(q+2)=4(1-4q)). For equal roots (D=0), so \(q=\frac{1}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(q=\frac{1}{4}\). Here (D=4(q-1)2 -4q(q+2)=4(1-4q)). For equal roots (D=0), so \(q=\frac{1}{4}\).
Step 3
Exam Tip
यहाँ (D=4(q-1)2 -4q(q+2)=4(1-4q)) है। समान मूलों के लिए (D=0), इसलिए \(q=\frac{1}{4}\)।
Login to save your score, XP, coins and progress. Login
समीकरण ((q+2)x-2 -2(q-1)x+q=0) में \(q\neq-2\) हो, तो समान मूलों के लिए (q) का मान क्या है?
In ((q+2)x-2 -2(q-1)x+q=0), with \(q\neq-2\), what is the value of (q) for equal roots?
#quadratic-equations
#equal-roots
#parameter
A \(\frac{1}{2}\)
B (2)
C \(-\frac{1}{2}\)
D (0)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{2}\)
Step 1
Concept
Here (D=4(q-1)2 -4q(q+2)=4(1-4q)). Setting (D=0) gives \(q=\frac{1}{4}\), so calculate carefully.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). Here (D=4(q-1)2 -4q(q+2)=4(1-4q)). Setting (D=0) gives \(q=\frac{1}{4}\), so calculate carefully.
Step 3
Exam Tip
यहाँ (D=4(q-1)2 -4q(q+2)=4(1-4q)) है। (D=0) से \(q=\frac{1}{4}\) नहीं, सही गणना जाँचें।
Login to save your score, XP, coins and progress. Login
समीकरण (x-2 -(s+4)x+4s=0) में समान मूलों के लिए (s) का मान क्या होगा?
What is the value of (s) for equal roots in (x-2 -(s+4)x+4s=0)?
#quadratic-equations
#equal-roots
#parameter
A (s=4)
B (s=-4)
C (s=0)
D (s=8)
Explanation opens after your attempt
Step 1
Concept
Here (D=(s+4)2 -16s=(s-4 )2 ). For equal roots (D=0), so (s=4).
Step 2
Why this answer is correct
The correct answer is A. (s=4). Here (D=(s+4)2 -16s=(s-4 )2 ). For equal roots (D=0), so (s=4).
Step 3
Exam Tip
यहाँ (D=(s+4)2 -16s=(s-4 )2 ) है। समान मूलों के लिए (D=0), इसलिए (s=4)।
Login to save your score, XP, coins and progress. Login
यदि किसी द्विघात का विविक्तकर \(D=25-4q^2\) है, तो समान मूलों के लिए (q) के मान क्या होंगे?
If a quadratic has discriminant \(D=25-4q^2\), what are the values of (q) for equal roots?
#quadratic-equations
#discriminant-form
#equal-roots
A \(q=\frac{5}{2}\) या \(q=-\frac{5}{2}\) / \(q=\frac{5}{2}\) or \(q=-\frac{5}{2}\)
B (q=5) या (q=-5) / (q=5) or (q=-5)
C (q=2) या (q=-2) / (q=2) or (q=-2)
D (q=0)
Explanation opens after your attempt
Correct Answer
A. \(q=\frac{5}{2}\) या \(q=-\frac{5}{2}\) / \(q=\frac{5}{2}\) or \(q=-\frac{5}{2}\)
Step 1
Concept
For equal roots \(25-4q^2=0\) is needed. This gives \(q^2=\frac{25}{4}\), hence \(q=\pm\frac{5}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(q=\frac{5}{2}\) या \(q=-\frac{5}{2}\) / \(q=\frac{5}{2}\) or \(q=-\frac{5}{2}\). For equal roots \(25-4q^2=0\) is needed. This gives \(q^2=\frac{25}{4}\), hence \(q=\pm\frac{5}{2}\).
Step 3
Exam Tip
समान मूलों के लिए \(25-4q^2=0\) होना चाहिए। इससे \(q^2=\frac{25}{4}\), अतः \(q=\pm\frac{5}{2}\)।
Login to save your score, XP, coins and progress. Login
यदि (D=(s-2 )2 ) है, तो समान मूलों के लिए (s) का मान क्या होगा?
If (D=(s-2 )2 ), what is the value of (s) for equal roots?
#quadratic-equations
#discriminant-form
#equal-roots
A (s=2)
B (s=-2)
C (s=0)
D हर (s) / Every (s)
Explanation opens after your attempt
Step 1
Concept
Equal roots need (D=0). From ((s-2 )2 =0), we get (s=2).
Step 2
Why this answer is correct
The correct answer is A. (s=2). Equal roots need (D=0). From ((s-2 )2 =0), we get (s=2).
Step 3
Exam Tip
समान मूलों के लिए (D=0) चाहिए। ((s-2 )2 =0) से (s=2) मिलता है।
Login to save your score, XP, coins and progress. Login
निम्न में से किस समीकरण के मूल समान हैं?
Which of the following equations has equal roots?
#quadratic-equations
#choose-equation
#equal-roots
A \(7x^2-10\sqrt{7}x+25=0\)
B \(7x^2-9\sqrt{7}x+25=0\)
C \(7x^2-8\sqrt{7}x+25=0\)
D \(7x^2-6\sqrt{7}x+25=0\)
Explanation opens after your attempt
Correct Answer
A. \(7x^2-10\sqrt{7}x+25=0\)
Step 1
Concept
In option (A), (D=\(-10\sqrt{7}\)2 -4(7)(25)=0). Equal roots need (D=0).
Step 2
Why this answer is correct
The correct answer is A. \(7x^2-10\sqrt{7}x+25=0\). In option (A), (D=\(-10\sqrt{7}\)2 -4(7)(25)=0). Equal roots need (D=0).
Step 3
Exam Tip
विकल्प (A) में (D=\(-10\sqrt{7}\)2 -4(7)(25)=0) है। समान मूलों के लिए (D=0) चाहिए।
Login to save your score, XP, coins and progress. Login
समीकरण (3x-2 -2(2k+1)x+(k+1)2 =0) में समान मूलों के लिए (k) के मान क्या होंगे?
For equal roots in (3x-2 -2(2k+1)x+(k+1)2 =0), what are the values of (k)?
#quadratic-equations
#equal-roots
#parameter
A (k=1) या (k=-2) / (k=1) or (k=-2)
B (k=2) या (k=-1) / (k=2) or (k=-1)
C (k=0) या (k=-3) / (k=0) or (k=-3)
D (k=3) या (k=-1) / (k=3) or (k=-1)
Explanation opens after your attempt
Correct Answer
A. (k=1) या (k=-2) / (k=1) or (k=-2)
Step 1
Concept
Here (D=4(2k+1)2 -12(k+1)2 =4(k-1)(k+2)). From (D=0), (k=1) or (k=-2).
Step 2
Why this answer is correct
The correct answer is A. (k=1) या (k=-2) / (k=1) or (k=-2). Here (D=4(2k+1)2 -12(k+1)2 =4(k-1)(k+2)). From (D=0), (k=1) or (k=-2).
Step 3
Exam Tip
यहाँ (D=4(2k+1)2 -12(k+1)2 =4(k-1)(k+2)) है। (D=0) से (k=1) या (k=-2)।
Login to save your score, XP, coins and progress. Login
यदि (x-2 +2(m-2 )x+\(m^2-3m+4\)=0) के मूल समान हों, तो (m) का मान क्या होगा?
If (x-2 +2(m-2 )x+\(m^2-3m+4\)=0) has equal roots, what is the value of (m)?
#quadratic-equations
#equal-roots
#parameter
A (0)
B (1)
C (2)
D (4)
Explanation opens after your attempt
Step 1
Concept
Here (D=4(m-2 )2 -4\(m^2-3m+4\)=-4m). From (D=0), (m=0).
Step 2
Why this answer is correct
The correct answer is A. (0). Here (D=4(m-2 )2 -4\(m^2-3m+4\)=-4m). From (D=0), (m=0).
Step 3
Exam Tip
यहाँ (D=4(m-2 )2 -4\(m^2-3m+4\)=-4m) है। (D=0) से (m=0) मिलता है।
Login to save your score, XP, coins and progress. Login
समीकरण \(x^2-2\lambda x+\lambda=0\) के समान मूलों के लिए \(\lambda\) के मान कौन से हैं?
What are the values of \(\lambda\) for equal roots of \(x^2-2\lambda x+\lambda=0\)?
#quadratic-equations
#equal-roots
#parameter
A \(\lambda=0\) या \(\lambda=1\) / \(\lambda=0\) or \(\lambda=1\)
B केवल \(\lambda=1\) / Only \(\lambda=1\)
C केवल \(\lambda=0\) / Only \(\lambda=0\)
D \(\lambda=-1\) या \(\lambda=1\) / \(\lambda=-1\) or \(\lambda=1\)
Explanation opens after your attempt
Correct Answer
A. \(\lambda=0\) या \(\lambda=1\) / \(\lambda=0\) or \(\lambda=1\)
Step 1
Concept
For equal roots (D=4\lambda\(\lambda-1\)=0). Therefore \(\lambda=0\) or \(\lambda=1\).
Step 2
Why this answer is correct
The correct answer is A. \(\lambda=0\) या \(\lambda=1\) / \(\lambda=0\) or \(\lambda=1\). For equal roots (D=4\lambda\(\lambda-1\)=0). Therefore \(\lambda=0\) or \(\lambda=1\).
Step 3
Exam Tip
समान मूलों के लिए (D=4\lambda\(\lambda-1\)=0) है। इसलिए \(\lambda=0\) या \(\lambda=1\)।
Login to save your score, XP, coins and progress. Login
यदि समीकरण (kx-2 -2(k+1)x+(k+3)=0) में \(k\neq0\) हो, तो समान मूलों के लिए (k) का मान क्या होगा?
If \(k\neq0\) in (kx-2 -2(k+1)x+(k+3)=0), what is the value of (k) for equal roots?
#quadratic-equations
#parameter
#equal-roots
A (k=1)
B (k=-1)
C (k=3)
D (k=-3)
Explanation opens after your attempt
Step 1
Concept
For equal roots (D=4(k+1)2 -4k(k+3)=0). Simplifying gives (1-k=0), so (k=1).
Step 2
Why this answer is correct
The correct answer is A. (k=1). For equal roots (D=4(k+1)2 -4k(k+3)=0). Simplifying gives (1-k=0), so (k=1).
Step 3
Exam Tip
समान मूलों के लिए (D=4(k+1)2 -4k(k+3)=0) है। सरल करने पर (1-k=0), इसलिए (k=1)।
Login to save your score, XP, coins and progress. Login
समीकरण (2x-2 -(k+4)x+2k=0) के मूलों को समान बनाने वाला सही मान कौन सा है?
Which value makes the roots of (2x-2 -(k+4)x+2k=0) equal?
#quadratic-equations
#parameter
#equal-roots
A (k=4)
B (k=12)
C (k=-4)
D (k=0)
Explanation opens after your attempt
Step 1
Concept
Here (D=(k+4)2 -16k=(k-4)2 ). (D=0) only when (k=4).
Step 2
Why this answer is correct
The correct answer is A. (k=4). Here (D=(k+4)2 -16k=(k-4)2 ). (D=0) only when (k=4).
Step 3
Exam Tip
यहाँ (D=(k+4)2 -16k=(k-4)2 ) है। (D=0) केवल (k=4) पर होगा।
Login to save your score, XP, coins and progress. Login
समीकरण (2x-2 -(k+4)x+2k=0) के समान मूलों के लिए (k) के मान क्या होंगे?
What are the values of (k) for equal roots of (2x-2 -(k+4)x+2k=0)?
#quadratic-equations
#equal-roots
#parameter
A (k=4) या (k=12) / (k=4) or (k=12)
B (k=-4) या (k=-12) / (k=-4) or (k=-12)
C (k=2) या (k=8) / (k=2) or (k=8)
D (k=0) या (k=16) / (k=0) or (k=16)
Explanation opens after your attempt
Correct Answer
A. (k=4) या (k=12) / (k=4) or (k=12)
Step 1
Concept
For equal roots (D=(k+4)2 -16k=0). This gives \(k^2-8k+16=0\), so (k=4).
Step 2
Why this answer is correct
The correct answer is A. (k=4) या (k=12) / (k=4) or (k=12). For equal roots (D=(k+4)2 -16k=0). This gives \(k^2-8k+16=0\), so (k=4).
Step 3
Exam Tip
समान मूलों के लिए (D=(k+4)2 -16k=0) है। इससे \(k^2-8k+16=0\), इसलिए (k=4)।
Login to save your score, XP, coins and progress. Login
यदि \(px^2-18x+9=0\) के मूल समान हों और \(p\neq0\), तो (p) का मान क्या होगा?
If \(px^2-18x+9=0\) has equal roots and \(p\neq0\), what is the value of (p)?
#quadratic-equations
#parameter
#equal-roots
A (9)
B (18)
C (36)
D (81)
Explanation opens after your attempt
Step 1
Concept
Here (D=(-18)2 -4(p)(9)=0). This gives (324-36p=0), so (p=9).
Step 2
Why this answer is correct
The correct answer is A. (9). Here (D=(-18)2 -4(p)(9)=0). This gives (324-36p=0), so (p=9).
Step 3
Exam Tip
यहाँ (D=(-18)2 -4(p)(9)=0) है। इससे (324-36p=0), इसलिए (p=9)।
Login to save your score, XP, coins and progress. Login
समीकरण \(4x^2-9x+q=0\) के दो वास्तविक और समान मूलों के लिए (q) का मान क्या होगा?
What is the value of (q) for two real and equal roots of \(4x^2-9x+q=0\)?
#quadratic-equations
#parameter
#equal-roots
A \(q=\frac{81}{16}\)
B \(q=\frac{16}{81}\)
C \(q=\frac{9}{4}\)
D (q=81)
Explanation opens after your attempt
Correct Answer
A. \(q=\frac{81}{16}\)
Step 1
Concept
For equal roots (D=0). ((-9)2 -4(4)q=0) gives \(q=\frac{81}{16}\).
Step 2
Why this answer is correct
The correct answer is A. \(q=\frac{81}{16}\). For equal roots (D=0). ((-9)2 -4(4)q=0) gives \(q=\frac{81}{16}\).
Step 3
Exam Tip
समान मूलों के लिए (D=0) है। ((-9)2 -4(4)q=0) से \(q=\frac{81}{16}\)।
Login to save your score, XP, coins and progress. Login
समीकरण (x-2 -2(k-5)x+36=0) के समान मूलों के लिए (k) के मान क्या होंगे?
What are the values of (k) for equal roots of (x-2 -2(k-5)x+36=0)?
#quadratic-equations
#parameter
#equal-roots
A (k=11) या (k=-1) / (k=11) or (k=-1)
B (k=6) या (k=-6) / (k=6) or (k=-6)
C (k=5) या (k=-5) / (k=5) or (k=-5)
D (k=12) या (k=-2) / (k=12) or (k=-2)
Explanation opens after your attempt
Correct Answer
A. (k=11) या (k=-1) / (k=11) or (k=-1)
Step 1
Concept
For equal roots (4(k-5)2 -144=0). Thus \(k-5=\pm6\), so (k=11) or (k=-1).
Step 2
Why this answer is correct
The correct answer is A. (k=11) या (k=-1) / (k=11) or (k=-1). For equal roots (4(k-5)2 -144=0). Thus \(k-5=\pm6\), so (k=11) or (k=-1).
Step 3
Exam Tip
समान मूलों के लिए (4(k-5)2 -144=0) है। इससे \(k-5=\pm6\), इसलिए (k=11) या (k=-1)।
Login to save your score, XP, coins and progress. Login
कथन: \(7x^2-14x+7=0\) के मूल समान हैं। कारण: इसका (D=0) है। सही विकल्प चुनिए।
Assertion: \(7x^2-14x+7=0\) has equal roots. Reason: Its (D=0). Choose the correct option.
#quadratic-equations
#assertion-reason
#equal-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=(-14)2 -4(7)(7)=0). So the reason correctly explains the assertion.
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=(-14)2 -4(7)(7)=0). So the reason correctly explains the assertion.
Step 3
Exam Tip
यहाँ (D=(-14)2 -4(7)(7)=0) है। इसलिए कारण कथन को सही रूप से समझाता है।
Login to save your score, XP, coins and progress. Login
समीकरण (x-2 -2(k+4)x+25=0) में समान मूलों के लिए (k) के मान क्या होंगे?
For equal roots in (x-2 -2(k+4)x+25=0), what are the values of (k)?
#quadratic-equations
#parameter
#equal-roots
A (k=1) या (k=-9) / (k=1) or (k=-9)
B (k=5) या (k=-5) / (k=5) or (k=-5)
C (k=4) या (k=-4) / (k=4) or (k=-4)
D (k=9) या (k=-1) / (k=9) or (k=-1)
Explanation opens after your attempt
Correct Answer
A. (k=1) या (k=-9) / (k=1) or (k=-9)
Step 1
Concept
For equal roots (4(k+4)2 -100=0). Thus \(k+4=\pm5\), so (k=1) or (k=-9).
Step 2
Why this answer is correct
The correct answer is A. (k=1) या (k=-9) / (k=1) or (k=-9). For equal roots (4(k+4)2 -100=0). Thus \(k+4=\pm5\), so (k=1) or (k=-9).
Step 3
Exam Tip
समान मूलों के लिए (4(k+4)2 -100=0) है। इससे \(k+4=\pm5\), इसलिए (k=1) या (k=-9)।
Login to save your score, XP, coins and progress. Login
समीकरण \(kx^2+10x+5=0\) के मूल समान हों और \(k\neq0\), तो (k) का मान क्या होगा?
If \(kx^2+10x+5=0\) has equal roots and \(k\neq0\), what is the value of (k)?
#quadratic-equations
#parameter
#equal-roots
A (5)
B (10)
C (20)
D (25)
Explanation opens after your attempt
Step 1
Concept
Here (D=102 -4(k)(5)=0). This gives (100-20k=0), so (k=5).
Step 2
Why this answer is correct
The correct answer is A. (5). Here (D=102 -4(k)(5)=0). This gives (100-20k=0), so (k=5).
Step 3
Exam Tip
यहाँ (D=102 -4(k)(5)=0) है। इससे (100-20k=0), इसलिए (k=5)।
Login to save your score, XP, coins and progress. Login
