यदि \(x^2+kx+9=0\) के मूल बराबर हों, तो (k) के कौन से मान संभव हैं?
If \(x^2+kx+9=0\) has equal roots, which values of (k) are possible?
#quadratic-equations
#parameter
#equal-roots
A (k=6) या (k=-6) / (k=6) or (k=-6)
B केवल (k=6) / Only (k=6)
C केवल (k=-6) / Only (k=-6)
D (k=3) या (k=-3) / (k=3) or (k=-3)
Explanation opens after your attempt
Correct Answer
A. (k=6) या (k=-6) / (k=6) or (k=-6)
Step 1
Concept
Here \(D=k^2-36=0\) gives \(k^2=36\). Check both signs in square equations.
Step 2
Why this answer is correct
The correct answer is A. (k=6) या (k=-6) / (k=6) or (k=-6). Here \(D=k^2-36=0\) gives \(k^2=36\). Check both signs in square equations.
Step 3
Exam Tip
यहाँ \(D=k^2-36=0\) से \(k^2=36\) मिलता है। वर्ग समीकरण में दोनों चिन्हों को जाँचें।
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यदि \(x^2-4x+k=0\) के मूल समान हों, तो (k) का मान क्या होगा?
If \(x^2-4x+k=0\) has equal roots, what is the value of (k)?
#quadratic-equations
#equal-roots
#parameter
A (4)
B (-4)
C (8)
D (16)
Explanation opens after your attempt
Step 1
Concept
For equal roots (D=0), so ((-4)2 -4(1)(k)=0) gives (k=4). Use (D=0) in such questions.
Step 2
Why this answer is correct
The correct answer is A. (4). For equal roots (D=0), so ((-4)2 -4(1)(k)=0) gives (k=4). Use (D=0) in such questions.
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए ((-4)2 -4(1)(k)=0) से (k=4)। ऐसे प्रश्नों में (D=0) लगाएँ।
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यदि \(x^2+tx+12=0\) के वास्तविक मूल नहीं हैं तो (t) के लिए सही शर्त क्या है?
If \(x^2+tx+12=0\) has no real roots, what is the correct condition for (t)?
#quadratic equations
#parameter
#no real roots
A \(t^2<48\)
B \(t^2=48\)
C \(t^2>48\)
D \(t^2=12\)
Explanation opens after your attempt
Correct Answer
A. \(t^2<48\)
Step 1
Concept
For no real roots, (D<0) is required. So \(t^2-48<0\), that is \(t^2<48\).
Step 2
Why this answer is correct
The correct answer is A. \(t^2<48\). For no real roots, (D<0) is required. So \(t^2-48<0\), that is \(t^2<48\).
Step 3
Exam Tip
वास्तविक मूल नहीं होने के लिए (D<0) होना चाहिए। इसलिए \(t^2-48<0\), अर्थात \(t^2<48\)।
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यदि \(4x^2+sx+9=0\) के मूल वास्तविक और समान हैं तो \(s^2\) का मान क्या होगा?
If the roots of \(4x^2+sx+9=0\) are real and equal, what will be the value of \(s^2\)?
#quadratic equations
#parameter
#equal roots
A (144)
B (36)
C (72)
D (16)
Explanation opens after your attempt
Step 1
Concept
For real and equal roots (D=0). So (s-2 -4(4)(9)=0), hence \(s^2=144\).
Step 2
Why this answer is correct
The correct answer is A. (144). For real and equal roots (D=0). So (s-2 -4(4)(9)=0), hence \(s^2=144\).
Step 3
Exam Tip
वास्तविक और समान मूलों के लिए (D=0) होता है। इसलिए (s-2 -4(4)(9)=0), अतः \(s^2=144\)।
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यदि \(x^2+qx+10=0\) के वास्तविक मूल नहीं हैं तो (q) के लिए सही शर्त क्या है?
If \(x^2+qx+10=0\) has no real roots, what is the correct condition for (q)?
#quadratic equations
#parameter
#no real roots
A \(q^2<40\)
B \(q^2=40\)
C \(q^2>40\)
D \(q^2=10\)
Explanation opens after your attempt
Correct Answer
A. \(q^2<40\)
Step 1
Concept
For no real roots, (D<0) is needed. So \(q^2-40<0\), that is \(q^2<40\).
Step 2
Why this answer is correct
The correct answer is A. \(q^2<40\). For no real roots, (D<0) is needed. So \(q^2-40<0\), that is \(q^2<40\).
Step 3
Exam Tip
वास्तविक मूल नहीं होने के लिए (D<0) चाहिए। इसलिए \(q^2-40<0\), अर्थात \(q^2<40\)।
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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 \(x^2+kx+16=0\) के मूल समान हैं तो \(k^2\) का मान क्या है?
If the roots of \(x^2+kx+16=0\) are equal, what is the value of \(k^2\)?
#quadratic equations
#parameter
#equal roots
A (64)
B (16)
C (32)
D (8)
Explanation opens after your attempt
Step 1
Concept
For equal roots (D=0), so \(k^2-64=0\). Therefore \(k^2=64\).
Step 2
Why this answer is correct
The correct answer is A. (64). For equal roots (D=0), so \(k^2-64=0\). Therefore \(k^2=64\).
Step 3
Exam Tip
समान मूलों के लिए (D=0) होता है इसलिए \(k^2-64=0\)। अतः \(k^2=64\) है।
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यदि \(3x^2+mx+12=0\) के मूल वास्तविक और समान हैं तो \(m^2\) का मान क्या होगा?
If the roots of \(3x^2+mx+12=0\) are real and equal, what will be the value of \(m^2\)?
#quadratic equations
#equal roots
#parameter
A (144)
B (36)
C (72)
D (12)
Explanation opens after your attempt
Step 1
Concept
For equal roots (D=0), so (m-2 -4(3)(12)=0). Therefore \(m^2=144\).
Step 2
Why this answer is correct
The correct answer is A. (144). For equal roots (D=0), so (m-2 -4(3)(12)=0). Therefore \(m^2=144\).
Step 3
Exam Tip
समान मूलों के लिए (D=0) होता है इसलिए (m-2 -4(3)(12)=0)। अतः \(m^2=144\) है।
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यदि \(x^2+px+4=0\) के मूल वास्तविक और भिन्न हैं तो (p) के लिए सही शर्त क्या है?
If the roots of \(x^2+px+4=0\) are real and distinct, what is the correct condition for (p)?
#quadratic equations
#parameter
#real distinct roots
A \(p^2>16\)
B \(p^2=16\)
C \(p^2<16\)
D \(p^2=4\)
Explanation opens after your attempt
Correct Answer
A. \(p^2>16\)
Step 1
Concept
For real and distinct roots (D>0), so \(p^2-16>0\). Therefore \(p^2>16\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(p^2>16\). For real and distinct roots (D>0), so \(p^2-16>0\). Therefore \(p^2>16\) is correct.
Step 3
Exam Tip
वास्तविक और भिन्न मूलों के लिए (D>0) होता है इसलिए \(p^2-16>0\)। अतः \(p^2>16\) सही है।
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यदि \(x^2-2kx+16=0\) के मूल समान हैं तो \(k^2\) कितना होगा?
If the roots of \(x^2-2kx+16=0\) are equal, what will be \(k^2\)?
#quadratic equations
#parameter
#D zero
A (16)
B (8)
C (32)
D (64)
Explanation opens after your attempt
Step 1
Concept
For equal roots (D=0), so ((-2k)2 -4(1)(16)=0). This gives \(k^2=16\).
Step 2
Why this answer is correct
The correct answer is A. (16). For equal roots (D=0), so ((-2k)2 -4(1)(16)=0). This gives \(k^2=16\).
Step 3
Exam Tip
समान मूलों के लिए (D=0) है इसलिए ((-2k)2 -4(1)(16)=0)। इससे \(k^2=16\) मिलता है।
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यदि \(x^2+kx+9=0\) के मूल समान हों तो (k) का संभावित मान कौन सा है?
If the roots of \(x^2+kx+9=0\) are equal, which is a possible value of (k)?
#quadratic equations
#parameter
#equal roots
A (6)
B (3)
C (0)
D (9)
Explanation opens after your attempt
Step 1
Concept
From \(k^2=36\), we get (k=6) or (k=-6). Among the given options (6) is correct.
Step 2
Why this answer is correct
The correct answer is A. (6). From \(k^2=36\), we get (k=6) or (k=-6). Among the given options (6) is correct.
Step 3
Exam Tip
\(k^2=36\) से (k=6) या (k=-6) मिलता है। दिए गए विकल्पों में (6) सही है।
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यदि \(x^2+kx+9=0\) के मूल समान हैं तो (k) के लिए सही शर्त क्या है?
If the roots of \(x^2+kx+9=0\) are equal, what is the correct condition for (k)?
#quadratic equations
#equal roots
#parameter
A \(k^2=36\)
B \(k^2=9\)
C \(k^2=0\)
D \(k^2=18\)
Explanation opens after your attempt
Correct Answer
A. \(k^2=36\)
Step 1
Concept
For equal roots (D=0), so \(k^2-36=0\). Hence \(k^2=36\) is correct.
Step 2
Why this answer is correct
The correct answer is A. \(k^2=36\). For equal roots (D=0), so \(k^2-36=0\). Hence \(k^2=36\) is correct.
Step 3
Exam Tip
समान मूलों के लिए (D=0) होता है इसलिए \(k^2-36=0\)। अतः \(k^2=36\) सही है।
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समीकरण \(5x^2-4x+t=0\) के वास्तविक और समान मूलों के लिए (t) क्या होगा?
What is (t) for real and equal roots of \(5x^2-4x+t=0\)?
#quadratic equations
#parameter
#equal roots
#fraction
A \(\frac{4}{5}\)
B \(\frac{5}{4}\)
C (4)
D (5)
Explanation opens after your attempt
Correct Answer
A. \(\frac{4}{5}\)
Step 1
Concept
(D=(-4)2 -4(5)t=0) gives \(t=\frac{4}{5}\). Always set (D=0) for equal roots.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4}{5}\). (D=(-4)2 -4(5)t=0) gives \(t=\frac{4}{5}\). Always set (D=0) for equal roots.
Step 3
Exam Tip
(D=(-4)2 -4(5)t=0) से \(t=\frac{4}{5}\) मिलता है। समान मूलों के लिए हमेशा (D=0) रखें।
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समीकरण \(x^2-3x+s=0\) के समान वास्तविक मूलों के लिए (s) का मान क्या होगा?
What is the value of (s) for equal real roots of \(x^2-3x+s=0\)?
#quadratic equations
#parameter
#equal roots
#fraction
A \(\frac{9}{4}\)
B \(\frac{3}{4}\)
C (9)
D (3)
Explanation opens after your attempt
Correct Answer
A. \(\frac{9}{4}\)
Step 1
Concept
For equal roots (D=0), so ((-3)2 -4s=0) gives \(s=\frac{9}{4}\). Do not fear fractional answers.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9}{4}\). For equal roots (D=0), so ((-3)2 -4s=0) gives \(s=\frac{9}{4}\). Do not fear fractional answers.
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए ((-3)2 -4s=0) से \(s=\frac{9}{4}\)। भिन्न उत्तर से न डरें।
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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 \(3x^2+kx+12=0\) में समान वास्तविक मूलों के लिए (k) का कौन सा मान सही हो सकता है?
Which value of (k) can give equal real roots in \(3x^2+kx+12=0\)?
#quadratic equations
#parameter
#equal roots
A (12)
B (6)
C (9)
D (3)
Explanation opens after your attempt
Step 1
Concept
For equal roots (k-2 -4(3)(12)=0), so (k=12) or (k=-12). Among the options, (12) is correct.
Step 2
Why this answer is correct
The correct answer is A. (12). For equal roots (k-2 -4(3)(12)=0), so (k=12) or (k=-12). Among the options, (12) is correct.
Step 3
Exam Tip
समान मूलों के लिए (k-2 -4(3)(12)=0), इसलिए (k=12) या (k=-12)। दिए विकल्पों में (12) सही है।
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समीकरण \(x^2-10x+r=0\) में समान वास्तविक मूलों के लिए (r) का मान चुनिए।
Choose the value of (r) for equal real roots in \(x^2-10x+r=0\).
#quadratic equations
#parameter
#equal roots
A (25)
B (10)
C (20)
D (100)
Explanation opens after your attempt
Step 1
Concept
(D=(-10)2 -4r=0) gives (r=25). Equal roots have zero discriminant.
Step 2
Why this answer is correct
The correct answer is A. (25). (D=(-10)2 -4r=0) gives (r=25). Equal roots have zero discriminant.
Step 3
Exam Tip
(D=(-10)2 -4r=0) से (r=25) मिलता है। समान मूलों में विविक्तकर शून्य होता है।
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समीकरण \(x^2+4x+p=0\) के वास्तविक मूल न होने के लिए कौन सी शर्त सही है?
For \(x^2+4x+p=0\) to have no real roots, which condition is correct?
#quadratic equations
#parameter
#no real roots
#inequality
A (p>4)
B (p=4)
C (p<4)
D (p=0)
Explanation opens after your attempt
Step 1
Concept
For no real roots (D<0), so (16-4p<0) gives (p>4). A negative discriminant gives no real roots.
Step 2
Why this answer is correct
The correct answer is A. (p>4). For no real roots (D<0), so (16-4p<0) gives (p>4). A negative discriminant gives no real roots.
Step 3
Exam Tip
वास्तविक मूल न होने के लिए (D<0), इसलिए (16-4p<0) से (p>4)। ऋणात्मक विविक्तकर पर वास्तविक मूल नहीं होते।
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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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समीकरण \(x^2+kx+16=0\) के समान वास्तविक मूल होने के लिए (k) का कौन सा मान सही हो सकता है?
Which value of (k) can make \(x^2+kx+16=0\) have equal real roots?
#quadratic equations
#parameter
#equal roots
A (8)
B (4)
C (16)
D (2)
Explanation opens after your attempt
Step 1
Concept
For equal roots \(k^2-64=0\), so (k=8) or (k=-8). Among the options, (8) is correct.
Step 2
Why this answer is correct
The correct answer is A. (8). For equal roots \(k^2-64=0\), so (k=8) or (k=-8). Among the options, (8) is correct.
Step 3
Exam Tip
समान मूलों के लिए \(k^2-64=0\) होता है, इसलिए (k=8) या (k=-8)। दिए विकल्पों में (8) सही है।
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समीकरण \(x^2-8x+m=0\) में समान वास्तविक मूलों के लिए (m) क्या होगा?
In \(x^2-8x+m=0\), what is (m) for equal real roots?
#quadratic equations
#parameter
#equal roots
A (16)
B (8)
C (4)
D (64)
Explanation opens after your attempt
Step 1
Concept
(D=(-8)2 -4m=0) gives (m=16). Keep the discriminant zero for equal roots.
Step 2
Why this answer is correct
The correct answer is A. (16). (D=(-8)2 -4m=0) gives (m=16). Keep the discriminant zero for equal roots.
Step 3
Exam Tip
(D=(-8)2 -4m=0) से (m=16) मिलता है। बराबर मूलों के लिए विविक्तकर शून्य रखें।
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समीकरण \(x^2+6x+k=0\) के समान वास्तविक मूल होने के लिए (k) का मान क्या होगा?
For \(x^2+6x+k=0\) to have equal real roots, what should be the value of (k)?
#quadratic equations
#parameter
#equal roots
A (9)
B (6)
C (12)
D (36)
Explanation opens after your attempt
Step 1
Concept
For equal roots (D=0), so \(6^2-4k=0\) gives (k=9). Use (D=0) in such questions.
Step 2
Why this answer is correct
The correct answer is A. (9). For equal roots (D=0), so \(6^2-4k=0\) gives (k=9). Use (D=0) in such questions.
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए \(6^2-4k=0\) से (k=9)। ऐसे प्रश्न में (D=0) लगाएं।
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(x-2 -2(k+7)x+k-2 =0) के समान मूलों के लिए (k) क्या होगा?
What is (k) for equal roots of (x-2 -2(k+7)x+k-2 =0)?
#quadratic
#discriminant
#parameter
A \(k=-\frac{7}{2}\)
B \(k=\frac{7}{2}\)
C (k=-7)
D (k=7)
Explanation opens after your attempt
Correct Answer
A. \(k=-\frac{7}{2}\)
Step 1
Concept
(D=4(k+7)2 -4k-2 =0) gives ((k+7)2 =k-2 ), so (14k+49=0) and \(k=-\frac{7}{2}\). In exams, expand squares carefully.
Step 2
Why this answer is correct
The correct answer is A. \(k=-\frac{7}{2}\). (D=4(k+7)2 -4k-2 =0) gives ((k+7)2 =k-2 ), so (14k+49=0) and \(k=-\frac{7}{2}\). In exams, expand squares carefully.
Step 3
Exam Tip
(D=4(k+7)2 -4k-2 =0) से ((k+7)2 =k-2 ), इसलिए (14k+49=0) और \(k=-\frac{7}{2}\) है। परीक्षा में वर्ग फैलाते समय सावधानी रखें।
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यदि \(kx^2-26x+169=0\) के समान मूल हैं, तो (k) क्या होगा?
If \(kx^2-26x+169=0\) has equal roots, what is (k)?
#quadratic
#discriminant
#parameter
A (1)
B (2)
C (13)
D (169)
Explanation opens after your attempt
Step 1
Concept
For equal roots, (D=0), so (676-676k=0) and (k=1). In exams, keep (a=k) correctly.
Step 2
Why this answer is correct
The correct answer is A. (1). For equal roots, (D=0), so (676-676k=0) and (k=1). In exams, keep (a=k) correctly.
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए (676-676k=0) और (k=1) है। परीक्षा में (a=k) को ठीक से रखें।
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यदि \(x^2+px+64=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?
If one root of \(x^2+px+64=0\) is double the other and both are negative, what is (p)?
#quadratic
#roots-relation
#parameter
A \(12\sqrt{2}\)
B \(-12\sqrt{2}\)
C \(16\sqrt{2}\)
D (24)
Explanation opens after your attempt
Correct Answer
A. \(12\sqrt{2}\)
Step 1
Concept
Let the roots be (-r) and (-2r), then \(2r^2=64\) gives \(r=4\sqrt{2}\), and \(p=3r=12\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 2
Why this answer is correct
The correct answer is A. \(12\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=64\) gives \(r=4\sqrt{2}\), and \(p=3r=12\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=64\) से \(r=4\sqrt{2}\) और \(p=3r=12\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।
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यदि \(10x^2-23x+p=0\) के मूलों का गुणनफल \(\frac{2}{5}\) है, तो (p) क्या होगा?
If the product of roots of \(10x^2-23x+p=0\) is \(\frac{2}{5}\), what is (p)?
#quadratic
#product-of-roots
#parameter
A (4)
B (10)
C \( \frac{1}{25}\)
D \( \frac{23}{5}\)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{p}{10}\), so \(\frac{p}{10}=\frac{2}{5}\) gives (p=4). In exams, use the product formula \(\frac{c}{a}\).
Step 2
Why this answer is correct
The correct answer is A. (4). The product of roots is \(\frac{p}{10}\), so \(\frac{p}{10}=\frac{2}{5}\) gives (p=4). In exams, use the product formula \(\frac{c}{a}\).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{p}{10}\) है, इसलिए \(\frac{p}{10}=\frac{2}{5}\) से (p=4) है। परीक्षा में गुणनफल का सूत्र \(\frac{c}{a}\) लगाएं।
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यदि \(9x^2+px+36=0\) के मूलों का योग (-10) है, तो (p) क्या होगा?
If the sum of roots of \(9x^2+px+36=0\) is (-10), what is (p)?
#quadratic
#sum-of-roots
#parameter
A (90)
B (-90)
C (10)
D (324)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is \(-\frac{p}{9}\), so \(-\frac{p}{9}=-10\) gives (p=90). In exams, remember the sum formula \(-\frac{b}{a}\).
Step 2
Why this answer is correct
The correct answer is A. (90). The sum of roots is \(-\frac{p}{9}\), so \(-\frac{p}{9}=-10\) gives (p=90). In exams, remember the sum formula \(-\frac{b}{a}\).
Step 3
Exam Tip
मूलों का योग \(-\frac{p}{9}\) है, इसलिए \(-\frac{p}{9}=-10\) से (p=90) है। परीक्षा में योग का सूत्र \(-\frac{b}{a}\) याद रखें।
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यदि \(x^2-25x+q=0\) का एक मूल (10) है, तो (q) का मान क्या होगा?
If one root of \(x^2-25x+q=0\) is (10), what is the value of (q)?
#quadratic
#parameter
#product-of-roots
A (150)
B (250)
C (15)
D (25)
Explanation opens after your attempt
Step 1
Concept
The other root is (15), so \(q=10\times15=150\). In exams, when (a=1), the constant term is the product of roots.
Step 2
Why this answer is correct
The correct answer is A. (150). The other root is (15), so \(q=10\times15=150\). In exams, when (a=1), the constant term is the product of roots.
Step 3
Exam Tip
दूसरा मूल (15) है, इसलिए \(q=10\times15=150\) होगा। परीक्षा में (a=1) हो तो स्थिर पद मूलों का गुणनफल होता है।
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यदि (x-2 -2(k-5)x+k-2 -36=0) के मूल समान हों, तो (k) का मान क्या होगा?
If (x-2 -2(k-5)x+k-2 -36=0) has equal roots, what is the value of (k)?
#quadratic
#discriminant
#parameter
A \(k=\frac{61}{10}\)
B \(k=-\frac{61}{10}\)
C (k=5)
D (k=36)
Explanation opens after your attempt
Correct Answer
A. \(k=\frac{61}{10}\)
Step 1
Concept
For equal roots, (D=0), so ((k-5)2 =k-2 -36) and \(k=\frac{61}{10}\). In exams, handle constant terms carefully while expanding squares.
Step 2
Why this answer is correct
The correct answer is A. \(k=\frac{61}{10}\). For equal roots, (D=0), so ((k-5)2 =k-2 -36) and \(k=\frac{61}{10}\). In exams, handle constant terms carefully while expanding squares.
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए ((k-5)2 =k-2 -36) और \(k=\frac{61}{10}\) है। परीक्षा में वर्ग फैलाते समय स्थिर पद ध्यान से लें।
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(x-2 -2(k+6)x+k-2 =0) के समान मूलों के लिए (k) क्या होगा?
What is (k) for equal roots of (x-2 -2(k+6)x+k-2 =0)?
#quadratic
#discriminant
#parameter
A (k=-3)
B (k=3)
C (k=-6)
D (k=6)
Explanation opens after your attempt
Step 1
Concept
(D=4(k+6)2 -4k-2 =0) gives ((k+6)2 =k-2 ), so (12k+36=0) and (k=-3). In exams, expand squares carefully.
Step 2
Why this answer is correct
The correct answer is A. (k=-3). (D=4(k+6)2 -4k-2 =0) gives ((k+6)2 =k-2 ), so (12k+36=0) and (k=-3). In exams, expand squares carefully.
Step 3
Exam Tip
(D=4(k+6)2 -4k-2 =0) से ((k+6)2 =k-2 ), इसलिए (12k+36=0) और (k=-3) है। परीक्षा में वर्ग फैलाते समय सावधानी रखें।
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