Question 1/15
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 33
यदि \(x^2-11x+30=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{\alpha+1}{\alpha-1}+\frac{\beta+1}{\beta-1}\) का मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2-11x+30=0\), what is \(\frac{\alpha+1}{\alpha-1}+\frac{\beta+1}{\beta-1}\)?
Explanation opens after your attempt
Correct Answer
A. \(\frac{29}{10}\)
Step 1
Concept
The roots are (5) and (6). Hence \(\frac{6}{4}+\frac{7}{5}=\frac{15}{10}+\frac{14}{10}=\frac{29}{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{29}{10}\). The roots are (5) and (6). Hence \(\frac{6}{4}+\frac{7}{5}=\frac{15}{10}+\frac{14}{10}=\frac{29}{10}\).
Step 3
Exam Tip
जड़ें (5) और (6) हैं। इसलिए \(\frac{6}{4}+\frac{7}{5}=\frac{15}{10}+\frac{14}{10}=\frac{29}{10}\)।
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Question 2/15
Expert Mathematics
Quadratic Equations Roots of a Quadratic Equation Class 10 Level 32
यदि \(x^2-7x+12=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{\alpha+1}{\alpha-1}+\frac{\beta+1}{\beta-1}\) का मान क्या है?
If \(\alpha,\beta\) are the roots of \(x^2-7x+12=0\), what is \(\frac{\alpha+1}{\alpha-1}+\frac{\beta+1}{\beta-1}\)?
Explanation opens after your attempt
Correct Answer
C. \(\frac{11}{3}\)
Step 1
Concept
The roots are (3) and (4). Direct substitution gives \(\frac{4}{2}+\frac{5}{3}=\frac{11}{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{11}{3}\). The roots are (3) and (4). Direct substitution gives \(\frac{4}{2}+\frac{5}{3}=\frac{11}{3}\).
Step 3
Exam Tip
जड़ें (3) और (4) हैं। सीधे रखने पर \(\frac{4}{2}+\frac{5}{3}=\frac{11}{3}\) मिलता है।
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Question 3/15
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (x-2+(6k-3)x+5k=0) में (x=-1) मूल है, तो (k) का मान क्या है?
If (x=-1) is a root of (x-2+(6k-3)x+5k=0), what is the value of (k)?
Explanation opens after your attempt
Step 1
Concept
Putting (x=-1) gives (1-(6k-3)+5k=0). Thus (4-k=0), so (k=4).
Step 2
Why this answer is correct
The correct answer is A. (2). Putting (x=-1) gives (1-(6k-3)+5k=0). Thus (4-k=0), so (k=4).
Step 3
Exam Tip
(x=-1) रखने पर (1-(6k-3)+5k=0) मिलता है। इससे (4-k=0), इसलिए (k=4) होगा।
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Question 4/15
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (x=-6) समीकरण \(2x^2+px-18=0\) का मूल है, तो (p) का मान क्या होगा?
If (x=-6) is a root of \(2x^2+px-18=0\), what is the value of (p)?
Explanation opens after your attempt
Step 1
Concept
Putting (x=-6) gives (72-6p-18=0). Hence (p=9).
Step 2
Why this answer is correct
The correct answer is A. (9). Putting (x=-6) gives (72-6p-18=0). Hence (p=9).
Step 3
Exam Tip
(x=-6) रखने पर (72-6p-18=0) मिलता है। इससे (p=9) है।
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Question 5/15
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि (x-2+(5k-2)x+4k=0) में (x=-2) मूल है, तो (k) का मान क्या है?
If (x=-2) is a root of (x-2+(5k-2)x+4k=0), what is the value of (k)?
Explanation opens after your attempt
Step 1
Concept
Putting (x=-2) gives (4-2(5k-2)+4k=0). Thus (8-6k=0), so \(k=\frac{4}{3}\).
Step 2
Why this answer is correct
The correct answer is A. (0). Putting (x=-2) gives (4-2(5k-2)+4k=0). Thus (8-6k=0), so \(k=\frac{4}{3}\).
Step 3
Exam Tip
(x=-2) रखने पर (4-2(5k-2)+4k=0) मिलता है। इससे (8-6k=0), इसलिए \(k=\frac{4}{3}\)।
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Question 6/15
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि (x=-5) समीकरण \(3x^2+px-20=0\) का मूल है, तो (p) का मान क्या होगा?
If (x=-5) is a root of \(3x^2+px-20=0\), what is the value of (p)?
Explanation opens after your attempt
Step 1
Concept
Putting (x=-5) gives (75-5p-20=0). Hence (p=11).
Step 2
Why this answer is correct
The correct answer is A. (11). Putting (x=-5) gives (75-5p-20=0). Hence (p=11).
Step 3
Exam Tip
(x=-5) रखने पर (75-5p-20=0) मिलता है। इससे (p=11) है।
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Question 7/15
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
यदि (x-2+(4k-1)x+3k=0) में (x=-1) मूल है, तो (k) का मान क्या है?
If (x=-1) is a root of (x-2+(4k-1)x+3k=0), what is the value of (k)?
Explanation opens after your attempt
Step 1
Concept
Putting (x=-1) gives (1-(4k-1)+3k=0). Thus (2-k=0), so (k=2).
Step 2
Why this answer is correct
The correct answer is A. (0). Putting (x=-1) gives (1-(4k-1)+3k=0). Thus (2-k=0), so (k=2).
Step 3
Exam Tip
(x=-1) रखने पर (1-(4k-1)+3k=0) मिलता है। इससे (2-k=0) नहीं बल्कि (k=2) मिलता है।
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Question 8/15
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
यदि (x=-3) समीकरण \(4x^2+px-9=0\) का मूल है, तो (p) का मान क्या होगा?
If (x=-3) is a root of \(4x^2+px-9=0\), what is the value of (p)?
Explanation opens after your attempt
Step 1
Concept
Putting (x=-3) gives (36-3p-9=0). Hence (p=9).
Step 2
Why this answer is correct
The correct answer is A. (9). Putting (x=-3) gives (36-3p-9=0). Hence (p=9).
Step 3
Exam Tip
(x=-3) रखने पर (36-3p-9=0) मिलता है। इससे (p=9) है।
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Question 9/15
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (x-2+(3k+1)x+2k=0) में (x=-2) मूल है, तो (k) का मान क्या है?
If (x=-2) is a root of (x-2+(3k+1)x+2k=0), what is the value of (k)?
Explanation opens after your attempt
Correct Answer
A. \( \frac{1}{2} \)
Step 1
Concept
Putting (x=-2) gives (4-2(3k+1)+2k=0). Thus (2-4k=0), so \(k=\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{1}{2} \). Putting (x=-2) gives (4-2(3k+1)+2k=0). Thus (2-4k=0), so \(k=\frac{1}{2}\).
Step 3
Exam Tip
(x=-2) रखने पर (4-2(3k+1)+2k=0) मिलता है। इससे (2-4k=0), इसलिए \(k=\frac{1}{2}\)।
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Question 10/15
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (x=-4) समीकरण \(2x^2+px-12=0\) का मूल है, तो (p) का मान क्या होगा?
If (x=-4) is a root of \(2x^2+px-12=0\), what is the value of (p)?
Explanation opens after your attempt
Step 1
Concept
Putting (x=-4) gives (32-4p-12=0). Hence (p=5).
Step 2
Why this answer is correct
The correct answer is A. (5). Putting (x=-4) gives (32-4p-12=0). Hence (p=5).
Step 3
Exam Tip
(x=-4) रखने पर (32-4p-12=0) मिलता है। इससे (p=5) है।
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Question 11/15
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि (x-2+(2k-1)x+k=0) में (x=1) मूल है, तो (k) का मान क्या है?
If (x=1) is a root of (x-2+(2k-1)x+k=0), what is the value of (k)?
Explanation opens after your attempt
Step 1
Concept
Putting (x=1) gives (1+2k-1+k=0). Thus (3k=0), so (k=0).
Step 2
Why this answer is correct
The correct answer is A. (0). Putting (x=1) gives (1+2k-1+k=0). Thus (3k=0), so (k=0).
Step 3
Exam Tip
(x=1) रखने पर (1+2k-1+k=0) मिलता है। इससे (3k=0), इसलिए (k=0)।
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Question 12/15
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि (x=-2) समीकरण \(3x^2+px+10=0\) का मूल है, तो (p) का मान क्या होगा?
If (x=-2) is a root of \(3x^2+px+10=0\), what is the value of (p)?
Explanation opens after your attempt
Step 1
Concept
Putting (x=-2) gives (12-2p+10=0). Hence (p=11).
Step 2
Why this answer is correct
The correct answer is A. (11). Putting (x=-2) gives (12-2p+10=0). Hence (p=11).
Step 3
Exam Tip
(x=-2) रखने पर (12-2p+10=0) मिलता है। इससे (p=11) है।
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Question 13/15
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (x=4) समीकरण \(x^2+sx-32=0\) का मूल है, तो (s) का मान क्या होगा?
If (x=4) is a root of \(x^2+sx-32=0\), what is the value of (s)?
Explanation opens after your attempt
Step 1
Concept
Putting (x=4) gives (16+4s-32=0), so (s=4). When a root is given, substitute directly.
Step 2
Why this answer is correct
The correct answer is A. (4). Putting (x=4) gives (16+4s-32=0), so (s=4). When a root is given, substitute directly.
Step 3
Exam Tip
(x=4) रखने पर (16+4s-32=0) मिलता है, इसलिए (s=4) है। मूल दिया हो तो सीधा प्रतिस्थापन करें।
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Question 14/15
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि (x=-4) समीकरण \(x^2+tx-8=0\) का मूल है, तो (t) का मान क्या होगा?
If (x=-4) is a root of \(x^2+tx-8=0\), what is the value of (t)?
Explanation opens after your attempt
Step 1
Concept
Putting (x=-4) gives (16-4t-8=0), so (t=2). When a root is given, substitute it directly.
Step 2
Why this answer is correct
The correct answer is A. (2). Putting (x=-4) gives (16-4t-8=0), so (t=2). When a root is given, substitute it directly.
Step 3
Exam Tip
(x=-4) रखने पर (16-4t-8=0) मिलता है, इसलिए (t=2) है। मूल दिया हो तो सीधे प्रतिस्थापन करें।
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Question 15/15
Easy Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (x=2) समीकरण \(x^2+kx+2=0\) का मूल है तो (k) का मान क्या होगा?
If (x=2) is a root of \(x^2+kx+2=0\), what is the value of (k)?
Explanation opens after your attempt
Step 1
Concept
Putting (x=2) gives (4+2k+2=0), so (k=-3). If a root is given, substitute it in the equation.
Step 2
Why this answer is correct
The correct answer is B. (-3). Putting (x=2) gives (4+2k+2=0), so (k=-3). If a root is given, substitute it in the equation.
Step 3
Exam Tip
(x=2) रखने पर (4+2k+2=0) इसलिए (k=-3) मिलता है। मूल दिया हो तो उसे समीकरण में रखें।
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