Question 1/7
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (x=3) समीकरण \(kx^2-8x+5=0\) का मूल नहीं है, तो (k) पर कौन-सी शर्त होगी?
If (x=3) is not a root of \(kx^2-8x+5=0\), what condition must (k) satisfy?
#quadratic-equations
#not-root
#parameter
#expert
A \(k\neq \frac{19}{9}\)
B \(k=\frac{19}{9}\)
C \(k\neq8\)
D (k=8)
Explanation opens after your attempt
Correct Answer
A. \(k\neq \frac{19}{9}\)
Step 1
Concept
Putting (x=3), the left side becomes (9k-24+5=9k-19). For it not to be a root, \(9k-19\neq0\), so \(k\neq\frac{19}{9}\).
Step 2
Why this answer is correct
The correct answer is A. \(k\neq \frac{19}{9}\). Putting (x=3), the left side becomes (9k-24+5=9k-19). For it not to be a root, \(9k-19\neq0\), so \(k\neq\frac{19}{9}\).
Step 3
Exam Tip
(x=3) रखने पर बायां पक्ष (9k-24+5=9k-19) होता है। मूल न होने के लिए \(9k-19\neq0\), इसलिए \(k\neq\frac{19}{9}\)।
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Question 2/7
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि (x=-2) समीकरण \(kx^2+5x+6=0\) का मूल नहीं है, तो (k) पर कौन-सी शर्त होगी?
If (x=-2) is not a root of \(kx^2+5x+6=0\), what condition must (k) satisfy?
#quadratic-equations
#not-root
#parameter
#expert
A \(k\neq1\)
B (k=1)
C \(k\neq2\)
D (k=2)
Explanation opens after your attempt
Correct Answer
A. \(k\neq1\)
Step 1
Concept
Putting (x=-2), the left side becomes (4k-10+6=4k-4). For it not to be a root, \(4k-4\neq0\), so \(k\neq1\).
Step 2
Why this answer is correct
The correct answer is A. \(k\neq1\). Putting (x=-2), the left side becomes (4k-10+6=4k-4). For it not to be a root, \(4k-4\neq0\), so \(k\neq1\).
Step 3
Exam Tip
(x=-2) रखने पर बायां पक्ष (4k-10+6=4k-4) होता है। मूल न होने के लिए \(4k-4\neq0\), इसलिए \(k\neq1\)।
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Question 3/7
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
यदि (x=2) समीकरण \(kx^2-7x+3=0\) का मूल नहीं है, तो (k) पर कौन-सी शर्त होगी?
If (x=2) is not a root of \(kx^2-7x+3=0\), what condition must (k) satisfy?
#quadratic-equations
#not-root
#parameter
#expert
A \(k\neq \frac{11}{4}\)
B \(k=\frac{11}{4}\)
C \(k\neq7\)
D (k=7)
Explanation opens after your attempt
Correct Answer
A. \(k\neq \frac{11}{4}\)
Step 1
Concept
Putting (x=2), the left side becomes (4k-14+3=4k-11). For it not to be a root, \(4k-11\neq0\), so \(k\neq\frac{11}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(k\neq \frac{11}{4}\). Putting (x=2), the left side becomes (4k-14+3=4k-11). For it not to be a root, \(4k-11\neq0\), so \(k\neq\frac{11}{4}\).
Step 3
Exam Tip
(x=2) रखने पर बायां पक्ष (4k-14+3=4k-11) होता है। मूल न होने के लिए \(4k-11\neq0\), इसलिए \(k\neq\frac{11}{4}\)।
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Question 4/7
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (x=-1) समीकरण \(kx^2+3x+2=0\) का मूल नहीं है, तो (k) पर कौन-सी शर्त होगी?
If (x=-1) is not a root of \(kx^2+3x+2=0\), what condition must (k) satisfy?
#quadratic-equations
#not-root
#parameter
#hard
A \(k\neq1\)
B (k=1)
C \(k\neq3\)
D (k=3)
Explanation opens after your attempt
Correct Answer
A. \(k\neq1\)
Step 1
Concept
Putting (x=-1), the left side becomes (k-3+2=k-1). For it not to be a root, \(k-1\neq0\), so \(k\neq1\).
Step 2
Why this answer is correct
The correct answer is A. \(k\neq1\). Putting (x=-1), the left side becomes (k-3+2=k-1). For it not to be a root, \(k-1\neq0\), so \(k\neq1\).
Step 3
Exam Tip
(x=-1) रखने पर बायां पक्ष (k-3+2=k-1) होता है। मूल न होने के लिए \(k-1\neq0\), इसलिए \(k\neq1\)।
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Question 5/7
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि (x=1) समीकरण \(kx^2-5x+2=0\) का मूल नहीं है, तो (k) पर कौन-सी शर्त होगी?
If (x=1) is not a root of \(kx^2-5x+2=0\), what condition must (k) satisfy?
#quadratic-equations
#not-root
#parameter
#hard
A \(k\neq3\)
B (k=3)
C \(k\neq5\)
D (k=5)
Explanation opens after your attempt
Correct Answer
A. \(k\neq3\)
Step 1
Concept
Putting (x=1), the left side becomes (k-5+2=k-3). For it not to be a root, \(k-3\neq0\), so \(k\neq3\).
Step 2
Why this answer is correct
The correct answer is A. \(k\neq3\). Putting (x=1), the left side becomes (k-5+2=k-3). For it not to be a root, \(k-3\neq0\), so \(k\neq3\).
Step 3
Exam Tip
(x=1) रखने पर बायां पक्ष (k-5+2=k-3) होता है। मूल न होने के लिए \(k-3\neq0\), इसलिए \(k\neq3\)।
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Question 6/7
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
किस समीकरण में (x=-2) मूल नहीं है?
In which equation is (x=-2) not a root?
#quadratic-equations
#root-check
#not-root
#medium
A \(x^2+2x=0\)
B \(x^2+5x+6=0\)
C \(2x^2+3x-2=0\)
D \(x^2-2x+4=0\)
Explanation opens after your attempt
Correct Answer
D. \(x^2-2x+4=0\)
Step 1
Concept
Putting (x=-2) gives \(4+4+4=12\neq0\). To check a non-root, use substitution too.
Step 2
Why this answer is correct
The correct answer is D. \(x^2-2x+4=0\). Putting (x=-2) gives \(4+4+4=12\neq0\). To check a non-root, use substitution too.
Step 3
Exam Tip
(x=-2) रखने पर \(4+4+4=12\neq0\) मिलता है। मूल न होने की जांच भी प्रतिस्थापन से करें।
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Question 7/7
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
किस समीकरण में (x=1) मूल नहीं है?
In which equation is (x=1) not a root?
#quadratic-equations
#root-check
#not-root
#medium
A \(x^2-1=0\)
B \(x^2-3x+2=0\)
C \(2x^2+x-3=0\)
D \(x^2+x+1=0\)
Explanation opens after your attempt
Correct Answer
D. \(x^2+x+1=0\)
Step 1
Concept
Putting (x=1) gives \(1+1+1=3\neq 0\). To check when a value is not a root, use substitution too.
Step 2
Why this answer is correct
The correct answer is D. \(x^2+x+1=0\). Putting (x=1) gives \(1+1+1=3\neq 0\). To check when a value is not a root, use substitution too.
Step 3
Exam Tip
(x=1) रखने पर \(1+1+1=3\neq 0\) मिलता है। किसी विकल्प में मूल न होने की जांच भी प्रतिस्थापन से करें।
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