Question 31/92
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
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 \(a=-\frac{7}{10}\)।
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Question 32/92
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
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)।
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Question 33/92
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
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)।
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Question 34/92
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
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।
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Question 35/92
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
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।
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Question 36/92
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
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)।
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Question 37/92
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
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 \(k=\frac{1}{4}\)।
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Question 38/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
((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।
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Question 39/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
\(4x^2-4x+1=0\) की दोनों जड़ों का मान क्या है?
What is the value of both roots of \(4x^2-4x+1=0\)?
#quadratic-roots
#equal-roots
#perfect-square
A \(\frac{1}{2}\)
B \(-\frac{1}{2}\)
C (1)
D (-1)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{2}\)
Step 1
Concept
(4x-2 -4x+1=(2x-1)2 ). Therefore the equal root is \(x=\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). (4x-2 -4x+1=(2x-1)2 ). Therefore the equal root is \(x=\frac{1}{2}\).
Step 3
Exam Tip
(4x-2 -4x+1=(2x-1)2 ) है। इसलिए समान जड़ \(x=\frac{1}{2}\) है।
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Question 40/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
\(x^2+kx+16=0\) 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।
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Question 41/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
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।
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Question 42/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
\(2x^2+\lambda x+8=0\) 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।
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Question 43/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
(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)।
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Question 44/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
(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।
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Question 45/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि समीकरण (x-2 -2(m+2)x+(m+2)2 =0) के मूल बराबर हैं तो (m) के लिए क्या सही है?
If roots of (x-2 -2(m+2)x+(m+2)2 =0) are equal, what is true for (m)?
#roots
#equal_roots
#identity_parameter
A हर वास्तविक (m) / Every real (m)
B केवल (m=0) / Only (m=0)
C केवल (m=-2) / Only (m=-2)
D कोई वास्तविक (m) नहीं / No real (m)
Explanation opens after your attempt
Correct Answer
A. हर वास्तविक (m) / Every real (m)
Step 1
Concept
The equation becomes ((x-(m+2))2 =0). Therefore the roots are equal for every real (m).
Step 2
Why this answer is correct
The correct answer is A. हर वास्तविक (m) / Every real (m). The equation becomes ((x-(m+2))2 =0). Therefore the roots are equal for every real (m).
Step 3
Exam Tip
समीकरण ((x-(m+2))2 =0) बनता है। इसलिए हर वास्तविक (m) के लिए दोनों मूल बराबर होंगे।
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Question 46/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x^2-12x+36=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(\alpha-\beta\) का मान क्या है?
If \(\alpha\) and \(\beta\) are roots of \(x^2-12x+36=0\), what is the value of \(\alpha-\beta\)?
#roots
#equal_roots
#difference
A (0)
B (6)
C (12)
D (36)
Explanation opens after your attempt
Step 1
Concept
This is ((x-6)2 =0), so both roots are (6) and (6). Hence \(\alpha-\beta=0\).
Step 2
Why this answer is correct
The correct answer is A. (0). This is ((x-6)2 =0), so both roots are (6) and (6). Hence \(\alpha-\beta=0\).
Step 3
Exam Tip
यह ((x-6)2 =0) है इसलिए दोनों मूल (6) और (6) हैं। अतः \(\alpha-\beta=0\) है।
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Question 47/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि समीकरण \(x^2-6x+k=0\) के मूल वास्तविक और बराबर हैं तो (k) का मान क्या है?
If roots of \(x^2-6x+k=0\) are real and equal, what is the value of (k)?
#roots
#equal_roots
#discriminant
A (9)
B (6)
C (18)
D (36)
Explanation opens after your attempt
Step 1
Concept
For equal roots (D=0). From (36-4k=0), we get (k=9).
Step 2
Why this answer is correct
The correct answer is A. (9). For equal roots (D=0). From (36-4k=0), we get (k=9).
Step 3
Exam Tip
बराबर मूलों के लिए (D=0) होता है। (36-4k=0) से (k=9) मिलता है।
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Question 48/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि (x-2 -2(m+1)x+\(m^2+2m+1\)=0) के मूल बराबर हैं तो (m) के लिए क्या सही है?
If roots of (x-2 -2(m+1)x+\(m^2+2m+1\)=0) are equal, what is true for (m)?
#roots
#equal_roots
#identity_parameter
A हर वास्तविक (m) / Every real (m)
B केवल (m=0) / Only (m=0)
C केवल (m=-1) / Only (m=-1)
D कोई वास्तविक (m) नहीं / No real (m)
Explanation opens after your attempt
Correct Answer
A. हर वास्तविक (m) / Every real (m)
Step 1
Concept
The constant term is ((m+1)2 ), and the equation becomes ((x-(m+1))2 =0). Hence roots are equal for every real (m).
Step 2
Why this answer is correct
The correct answer is A. हर वास्तविक (m) / Every real (m). The constant term is ((m+1)2 ), and the equation becomes ((x-(m+1))2 =0). Hence roots are equal for every real (m).
Step 3
Exam Tip
अचर पद ((m+1)2 ) है और समीकरण ((x-(m+1))2 =0) बनता है। इसलिए हर वास्तविक (m) के लिए मूल बराबर हैं।
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Question 49/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि \(x^2-10x+25=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(\alpha-\beta\) का मान क्या है?
If \(\alpha\) and \(\beta\) are roots of \(x^2-10x+25=0\), what is the value of \(\alpha-\beta\)?
#roots
#equal_roots
#difference
A (0)
B (5)
C (10)
D (25)
Explanation opens after your attempt
Step 1
Concept
This is ((x-5)2 =0), so both roots are (5) and (5). Hence \(\alpha-\beta=0\).
Step 2
Why this answer is correct
The correct answer is A. (0). This is ((x-5)2 =0), so both roots are (5) and (5). Hence \(\alpha-\beta=0\).
Step 3
Exam Tip
यह ((x-5)2 =0) है इसलिए दोनों मूल (5) और (5) हैं। अतः \(\alpha-\beta=0\) है।
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Question 50/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
यदि समीकरण \(x^2-4x+k=0\) के मूल वास्तविक और बराबर हैं तो (k) का मान क्या है?
If roots of \(x^2-4x+k=0\) are real and equal, what is the value of (k)?
#roots
#equal_roots
#discriminant
A (4)
B (2)
C (8)
D (16)
Explanation opens after your attempt
Step 1
Concept
For equal roots (D=0). From (16-4k=0), we get (k=4).
Step 2
Why this answer is correct
The correct answer is A. (4). For equal roots (D=0). From (16-4k=0), we get (k=4).
Step 3
Exam Tip
बराबर मूलों के लिए (D=0) होता है। (16-4k=0) से (k=4) मिलता है।
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Question 51/92
Hard Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 31
समीकरण \(x^2-2kx+k^2=0\) के मूलों की प्रकृति क्या होगी?
What will be the nature of roots of \(x^2-2kx+k^2=0\)?
#roots
#equal_roots
#identity
A बराबर वास्तविक मूल / Equal real roots
B भिन्न वास्तविक मूल / Distinct real roots
C कोई वास्तविक मूल नहीं / No real roots
D केवल एक शून्य मूल / Only one zero root
Explanation opens after your attempt
Correct Answer
A. बराबर वास्तविक मूल / Equal real roots
Step 1
Concept
This equation becomes ((x-k)2 =0). Therefore both roots are (k) and (k), which are equal.
Step 2
Why this answer is correct
The correct answer is A. बराबर वास्तविक मूल / Equal real roots. This equation becomes ((x-k)2 =0). Therefore both roots are (k) and (k), which are equal.
Step 3
Exam Tip
यह समीकरण ((x-k)2 =0) बनता है। इसलिए दोनों मूल (k) और (k) बराबर हैं।
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Question 52/92
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि \(x^2+ax+25=0\) के मूल (5) और (5) हैं तो (a) का मान क्या है?
If the roots of \(x^2+ax+25=0\) are (5) and (5), what is the value of (a)?
#roots
#parameter
#equal_roots
A (-10)
B (10)
C (5)
D (-5)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (10), and (-a=10). Therefore (a=-10).
Step 2
Why this answer is correct
The correct answer is A. (-10). The sum of roots is (10), and (-a=10). Therefore (a=-10).
Step 3
Exam Tip
मूलों का योग (10) है और (-a=10) होगा। इसलिए (a=-10) है।
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Question 53/92
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि (D=0) और मूलों का योग (14) है तो प्रत्येक मूल क्या होगा?
If (D=0) and the sum of roots is (14), what will each root be?
#roots
#equal_roots
#sum
A (7)
B (14)
C (0)
D (28)
Explanation opens after your attempt
Step 1
Concept
When (D=0), both roots are equal. Therefore each root is \(\frac{14}{2}=7\).
Step 2
Why this answer is correct
The correct answer is A. (7). When (D=0), both roots are equal. Therefore each root is \(\frac{14}{2}=7\).
Step 3
Exam Tip
(D=0) होने पर दोनों मूल बराबर होते हैं। इसलिए प्रत्येक मूल \(\frac{14}{2}=7\) होगा।
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Question 54/92
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
समीकरण \(x^2+px+36=0\) के दोनों मूल (-6) और (-6) हैं तो (p) का मान क्या है?
If both roots of \(x^2+px+36=0\) are (-6) and (-6), what is the value of (p)?
#roots
#equal_roots
#parameter
A (12)
B (-12)
C (6)
D (-6)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (-12), and (-p=-12). Therefore (p=12).
Step 2
Why this answer is correct
The correct answer is A. (12). The sum of roots is (-12), and (-p=-12). Therefore (p=12).
Step 3
Exam Tip
मूलों का योग (-12) है और (-p=-12) होगा। इसलिए (p=12) है।
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Question 55/92
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
समीकरण \(x^2-10x+k=0\) के बराबर वास्तविक मूल होने के लिए (k) का मान क्या होगा?
For \(x^2-10x+k=0\) to have equal real roots, what should be the value of (k)?
#roots
#equal_roots
#parameter
A (25)
B (10)
C (5)
D (100)
Explanation opens after your attempt
Step 1
Concept
For equal roots (D=0). From (100-4k=0), we get (k=25).
Step 2
Why this answer is correct
The correct answer is A. (25). For equal roots (D=0). From (100-4k=0), we get (k=25).
Step 3
Exam Tip
बराबर मूलों के लिए (D=0) होता है। (100-4k=0) से (k=25) मिलता है।
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Question 56/92
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
समीकरण \(4x^2-20x+25=0\) का डिस्क्रिमिनेंट (D) क्या है?
What is the discriminant (D) of \(4x^2-20x+25=0\)?
#roots
#discriminant
#equal_roots
A (0)
B (20)
C (-20)
D (100)
Explanation opens after your attempt
Step 1
Concept
Here \(D=b^2-4ac=400-400=0\). Therefore its two real roots will be equal.
Step 2
Why this answer is correct
The correct answer is A. (0). Here \(D=b^2-4ac=400-400=0\). Therefore its two real roots will be equal.
Step 3
Exam Tip
यहां \(D=b^2-4ac=400-400=0\) है। इसलिए इसके दोनों वास्तविक मूल बराबर होंगे।
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Question 57/92
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x^2+ax+16=0\) के मूल (4) और (4) हैं तो (a) का मान क्या है?
If the roots of \(x^2+ax+16=0\) are (4) and (4), what is the value of (a)?
#roots
#parameter
#equal_roots
A (-8)
B (8)
C (4)
D (-4)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (8), and (-a=8). Therefore (a=-8).
Step 2
Why this answer is correct
The correct answer is A. (-8). The sum of roots is (8), and (-a=8). Therefore (a=-8).
Step 3
Exam Tip
मूलों का योग (8) है और (-a=8) होगा। इसलिए (a=-8) है।
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Question 58/92
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि (D=0) और मूलों का योग (12) है तो प्रत्येक मूल क्या होगा?
If (D=0) and the sum of roots is (12), what will each root be?
#roots
#equal_roots
#sum
A (6)
B (12)
C (0)
D (24)
Explanation opens after your attempt
Step 1
Concept
When (D=0), both roots are equal. Therefore each root is \(\frac{12}{2}=6\).
Step 2
Why this answer is correct
The correct answer is A. (6). When (D=0), both roots are equal. Therefore each root is \(\frac{12}{2}=6\).
Step 3
Exam Tip
(D=0) होने पर दोनों मूल बराबर होते हैं। इसलिए प्रत्येक मूल \(\frac{12}{2}=6\) होगा।
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Question 59/92
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
समीकरण \(x^2+px+25=0\) के दोनों मूल (-5) और (-5) हैं तो (p) का मान क्या है?
If both roots of \(x^2+px+25=0\) are (-5) and (-5), what is the value of (p)?
#roots
#equal_roots
#parameter
A (10)
B (-10)
C (5)
D (-5)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (-10), and (-p=-10). Therefore (p=10).
Step 2
Why this answer is correct
The correct answer is A. (10). The sum of roots is (-10), and (-p=-10). Therefore (p=10).
Step 3
Exam Tip
मूलों का योग (-10) है और (-p=-10) होगा। इसलिए (p=10) है।
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Question 60/92
Medium Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
समीकरण \(x^2-8x+k=0\) के बराबर वास्तविक मूल होने के लिए (k) का मान क्या होगा?
For \(x^2-8x+k=0\) to have equal real roots, what should be the value of (k)?
#roots
#equal_roots
#parameter
A (16)
B (8)
C (4)
D (64)
Explanation opens after your attempt
Step 1
Concept
For equal roots (D=0). From (64-4k=0), we get (k=16).
Step 2
Why this answer is correct
The correct answer is A. (16). For equal roots (D=0). From (64-4k=0), we get (k=16).
Step 3
Exam Tip
बराबर मूलों के लिए (D=0) होता है। (64-4k=0) से (k=16) मिलता है।
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