Question 241/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (4) और (5) समीकरण \(x^2-sx+p=0\) के मूल हैं, तो (s+p) का मान क्या है?
If (4) and (5) are roots of \(x^2-sx+p=0\), what is the value of (s+p)?
#quadratic-equations
#roots
#parameters
#hard
A (29)
B (20)
C (9)
D (25)
Explanation opens after your attempt
Step 1
Concept
The sum of roots gives (s=9) and the product gives (p=20). Therefore (s+p=29).
Step 2
Why this answer is correct
The correct answer is A. (29). The sum of roots gives (s=9) and the product gives (p=20). Therefore (s+p=29).
Step 3
Exam Tip
मूलों का योग (s=9) और गुणनफल (p=20) है। इसलिए (s+p=29) है।
Login to save your score, XP, coins and progress.
Question 242/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(3x^2-10x+7=0\) के मूलों के व्युत्क्रमों का योग क्या है?
What is the sum of reciprocals of the roots of \(3x^2-10x+7=0\)?
#quadratic-equations
#roots
#reciprocal-sum
#hard
A \( \frac{10}{7} \)
B \( \frac{7}{10} \)
C \( \frac{3}{7} \)
D \( \frac{10}{3} \)
Explanation opens after your attempt
Correct Answer
A. \( \frac{10}{7} \)
Step 1
Concept
The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{10}{3}}{\frac{7}{3}}=\frac{10}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{10}{7} \). The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{10}{3}}{\frac{7}{3}}=\frac{10}{7}\).
Step 3
Exam Tip
व्युत्क्रमों का योग \(\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ \(\frac{\frac{10}{3}}{\frac{7}{3}}=\frac{10}{7}\) है।
Login to save your score, XP, coins and progress.
Question 243/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(x^2+2kx+16=0\) के वास्तविक मूल होने की शर्त कौन-सी है?
What is the condition for \(x^2+2kx+16=0\) to have real roots?
#quadratic-equations
#real-roots
#parameter
#hard
A \(k\leq -4\) या \(k\geq 4\) / \(k\leq -4\) or \(k\geq 4\)
B (-4<k<4)
C (k=0)
D \(k\neq4\)
Explanation opens after your attempt
Correct Answer
A. \(k\leq -4\) या \(k\geq 4\) / \(k\leq -4\) or \(k\geq 4\)
Step 1
Concept
For real roots, \(D\geq0\) is needed. Here \(4k^2-64\geq0\), so \(k\leq-4\) or \(k\geq4\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq -4\) या \(k\geq 4\) / \(k\leq -4\) or \(k\geq 4\). For real roots, \(D\geq0\) is needed. Here \(4k^2-64\geq0\), so \(k\leq-4\) or \(k\geq4\).
Step 3
Exam Tip
वास्तविक मूलों के लिए \(D\geq0\) चाहिए। यहाँ \(4k^2-64\geq0\), इसलिए \(k\leq-4\) या \(k\geq4\)।
Login to save your score, XP, coins and progress.
Question 244/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2-2kx+25=0\) के मूल समान हैं, तो (k) के संभावित मान क्या हैं?
If the roots of \(x^2-2kx+25=0\) are equal, what are the possible values of (k)?
#quadratic-equations
#equal-roots
#discriminant
#hard
A \(k=\pm5\)
B \(k=\pm10\)
C (k=5)
D (k=-5)
Explanation opens after your attempt
Correct Answer
A. \(k=\pm5\)
Step 1
Concept
For equal roots, (D=0), so ((-2k)2 -100=0). This gives \(k^2=25\) and \(k=\pm5\).
Step 2
Why this answer is correct
The correct answer is A. \(k=\pm5\). For equal roots, (D=0), so ((-2k)2 -100=0). This gives \(k^2=25\) and \(k=\pm5\).
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए ((-2k)2 -100=0) मिलता है। इससे \(k^2=25\) और \(k=\pm5\) है।
Login to save your score, XP, coins and progress.
Question 245/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
मूलों का योग (-9) और गुणनफल (20) वाला मोनिक द्विघात समीकरण कौन-सा है?
Which monic quadratic equation has sum of roots (-9) and product (20)?
#quadratic-equations
#sum-product
#forming-equation
#hard
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\(=0). Substituting sum (-9) gives (x^2+9x+20=0).\)
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\(=0). Substituting sum (-9) gives (x^2+9x+20=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(योग (-9) रखने पर (x^2+9x+20=0) मिलता है\)।
Login to save your score, XP, coins and progress.
Question 246/600
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)?
#quadratic-equations
#root-substitution
#parameter
#hard
A (5)
B (-5)
C (11)
D (-11)
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।
Login to save your score, XP, coins and progress.
Question 247/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(\frac{x^2+1}{3}-\frac{x-2}{2}=5\) को पूर्णांक गुणांकों वाले मानक रूप में लिखिए।
Write \(\frac{x^2+1}{3}-\frac{x-2}{2}=5\) in standard form with integer coefficients.
#quadratic-equations
#fractions
#standard-form
#hard
A \(2x^2-3x-22=0\)
B \(2x^2-3x+22=0\)
C \(3x^2-2x-22=0\)
D \(2x^2+3x-22=0\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2-3x-22=0\)
Step 1
Concept
Multiplying the whole equation by (6) gives \(2x^2+2-3x+6=30\). Therefore the standard form is \(2x^2-3x-22=0\).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-3x-22=0\). Multiplying the whole equation by (6) gives \(2x^2+2-3x+6=30\). Therefore the standard form is \(2x^2-3x-22=0\).
Step 3
Exam Tip
पूरे समीकरण को (6) से गुणा करने पर \(2x^2+2-3x+6=30\) मिलता है। इसलिए मानक रूप \(2x^2-3x-22=0\) है।
Login to save your score, XP, coins and progress.
Question 248/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण ((2x-3)2 +(x+5)2 =34) का मानक रूप कौन-सा है?
What is the standard form of ((2x-3)2 +(x+5)2 =34)?
#quadratic-equations
#identity
#simplification
#hard
A \(5x^2-2x=0\)
B \(5x^2+2x=0\)
C \(5x^2-2x+34=0\)
D \(3x^2-2x=0\)
Explanation opens after your attempt
Correct Answer
A. \(5x^2-2x=0\)
Step 1
Concept
Expanding gives \(4x^2-12x+9+x^2+10x+25=34\). Simplifying gives \(5x^2-2x=0\).
Step 2
Why this answer is correct
The correct answer is A. \(5x^2-2x=0\). Expanding gives \(4x^2-12x+9+x^2+10x+25=34\). Simplifying gives \(5x^2-2x=0\).
Step 3
Exam Tip
विस्तार करने पर \(4x^2-12x+9+x^2+10x+25=34\) मिलता है। सरल करने पर \(5x^2-2x=0\) बनता है।
Login to save your score, XP, coins and progress.
Question 249/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (\(n^2-16\)x-2 -3x+7=0) द्विघात समीकरण है, तो (n) पर सही शर्त क्या है?
If (\(n^2-16\)x-2 -3x+7=0) is a quadratic equation, what is the correct condition on (n)?
#quadratic-equations
#parameter
#condition
#hard
A \(n\neq 4\)
B \(n\neq -4\)
C \(n\neq \pm4\)
D \(n=\pm4\)
Explanation opens after your attempt
Correct Answer
C. \(n\neq \pm4\)
Step 1
Concept
For the equation to be quadratic, \(n^2-16\neq0\) is needed. Hence both \(n\neq4\) and \(n\neq-4\) are necessary.
Step 2
Why this answer is correct
The correct answer is C. \(n\neq \pm4\). For the equation to be quadratic, \(n^2-16\neq0\) is needed. Hence both \(n\neq4\) and \(n\neq-4\) are necessary.
Step 3
Exam Tip
द्विघात होने के लिए \(n^2-16\neq0\) होना चाहिए। इसलिए \(n\neq4\) और \(n\neq-4\) दोनों जरूरी हैं।
Login to save your score, XP, coins and progress.
Question 250/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण ((6x+1)(x-4)=5x) का मानक द्विघात रूप कौन-सा है?
What is the standard quadratic form of ((6x+1)(x-4)=5x)?
#quadratic-equations
#standard-form
#expansion
#hard
A \(6x^2-28x-4=0\)
B \(6x^2-18x-4=0\)
C \(6x^2-23x-4=0\)
D \(6x^2+28x-4=0\)
Explanation opens after your attempt
Correct Answer
A. \(6x^2-28x-4=0\)
Step 1
Concept
Here ((6x+1)(x-4)=6x-2 -23x-4), and subtracting (5x) gives \(6x^2-28x-4=0\). First expand and then bring all terms to one side.
Step 2
Why this answer is correct
The correct answer is A. \(6x^2-28x-4=0\). Here ((6x+1)(x-4)=6x-2 -23x-4), and subtracting (5x) gives \(6x^2-28x-4=0\). First expand and then bring all terms to one side.
Step 3
Exam Tip
((6x+1)(x-4)=6x-2 -23x-4) है और (5x) घटाने पर \(6x^2-28x-4=0\) मिलता है। पहले विस्तार करें फिर सभी पद एक ओर लाएं।
Login to save your score, XP, coins and progress.
Question 251/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2-2ax+a^2-9=0\) है, तो मूलों का अंतर क्या होगा?
If \(x^2-2ax+a^2-9=0\), what will be the difference of the roots?
#quadratic-equations
#identity
#roots-difference
#hard
A (6)
B (3)
C (2a)
D (a+3)
Explanation opens after your attempt
Step 1
Concept
The equation is ((x-a)2 -9=0), so the roots are (a+3) and (a-3). Their difference is (6).
Step 2
Why this answer is correct
The correct answer is A. (6). The equation is ((x-a)2 -9=0), so the roots are (a+3) and (a-3). Their difference is (6).
Step 3
Exam Tip
समीकरण ((x-a)2 -9=0) है, इसलिए मूल (a+3) और (a-3) हैं। उनका अंतर (6) है।
Login to save your score, XP, coins and progress.
Question 252/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण \(x^2+4x+k=0\) के कोई वास्तविक मूल नहीं हैं। (k) पर सही शर्त क्या है?
The equation \(x^2+4x+k=0\) has no real roots. What is the correct condition on (k)?
#quadratic-equations
#no-real-roots
#parameter
#hard
A (k>4)
B (k=4)
C (k<4)
D \(k\leq4\)
Explanation opens after your attempt
Step 1
Concept
For no real roots, (D<0) is needed. Here (16-4k<0), so (k>4).
Step 2
Why this answer is correct
The correct answer is A. (k>4). For no real roots, (D<0) is needed. Here (16-4k<0), so (k>4).
Step 3
Exam Tip
कोई वास्तविक मूल न होने के लिए (D<0) चाहिए। यहाँ (16-4k<0), इसलिए (k>4)।
Login to save your score, XP, coins and progress.
Question 253/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2+3x-18=0\) के मूल \(\alpha,\beta\) हैं, तो \(\alpha\beta-\alpha-\beta\) का मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2+3x-18=0\), what is \(\alpha\beta-\alpha-\beta\)?
#quadratic-equations
#roots
#expression-value
#hard
A (-15)
B (-21)
C (15)
D (21)
Explanation opens after your attempt
Step 1
Concept
Here \(\alpha+\beta=-3\) and \(\alpha\beta=-18\). Thus (\alpha\beta-\alpha-\beta=-18-(-3)=-15).
Step 2
Why this answer is correct
The correct answer is A. (-15). Here \(\alpha+\beta=-3\) and \(\alpha\beta=-18\). Thus (\alpha\beta-\alpha-\beta=-18-(-3)=-15).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=-3\) और \(\alpha\beta=-18\) है। इसलिए (\alpha\beta-\alpha-\beta=-18-(-3)=-15)।
Login to save your score, XP, coins and progress.
Question 254/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2+2x-8=0\) के मूल \(\alpha,\beta\) हैं, तो (\(\alpha+\beta\)2 ) का मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2+2x-8=0\), what is (\(\alpha+\beta\)2 )?
#quadratic-equations
#roots
#sum-square
#hard
A (4)
B (8)
C (16)
D (64)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (-2). Therefore (\(\alpha+\beta\)2 =(-2)2 =4).
Step 2
Why this answer is correct
The correct answer is A. (4). The sum of roots is (-2). Therefore (\(\alpha+\beta\)2 =(-2)2 =4).
Step 3
Exam Tip
मूलों का योग (-2) है। इसलिए (\(\alpha+\beta\)2 =(-2)2 =4)।
Login to save your score, XP, coins and progress.
Question 255/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
एक समकोण त्रिभुज का आधार (x+1), ऊंचाई (x+3) और क्षेत्रफल (30) है। सही समीकरण कौन-सा है?
A right triangle has base (x+1), height (x+3), and area (30). Which equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#hard
A \(x^2+4x-57=0\)
B \(x^2+4x-30=0\)
C \(x^2+4x+3=0\)
D \(x^2+4x-60=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+4x-57=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+1)(x+3)=30). Thus ((x+1)(x+3)=60) and \(x^2+4x-57=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+4x-57=0\). The area is (\frac{1}{2}(x+1)(x+3)=30). Thus ((x+1)(x+3)=60) and \(x^2+4x-57=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+1)(x+3)=30) होगा। इसलिए ((x+1)(x+3)=60) और \(x^2+4x-57=0\)।
Login to save your score, XP, coins and progress.
Question 256/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
किस समीकरण के मूलों का गुणनफल ऋणात्मक होगा?
Which equation will have a negative product of roots?
#quadratic-equations
#product-of-roots
#sign
#hard
A \(2x^2+3x-5=0\)
B \(2x^2+3x+5=0\)
C \(2x^2-3x+5=0\)
D \(2x^2-3x+0=0\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2+3x-5=0\)
Step 1
Concept
The product of roots is \(\frac{c}{a}\). In the first option, \(\frac{-5}{2}<0\), so the product is negative.
Step 2
Why this answer is correct
The correct answer is A. \(2x^2+3x-5=0\). The product of roots is \(\frac{c}{a}\). In the first option, \(\frac{-5}{2}<0\), so the product is negative.
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{c}{a}\) है। पहले विकल्प में \(\frac{-5}{2}<0\), इसलिए गुणनफल ऋणात्मक है।
Login to save your score, XP, coins and progress.
Question 257/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2-10x+21=0\) के मूल \(\alpha,\beta\) हैं, तो (\(\alpha-3\)\(\beta-3\)) का मान क्या है?
If \(\alpha,\beta\) are roots of \(x^2-10x+21=0\), what is (\(\alpha-3\)\(\beta-3\))?
#quadratic-equations
#roots
#expression-value
#hard
A (0)
B (3)
C (21)
D (-9)
Explanation opens after your attempt
Step 1
Concept
(\(\alpha-3\)\(\beta-3\)=\alpha\beta-3\(\alpha+\beta\)+9). Here (21-30+9=0).
Step 2
Why this answer is correct
The correct answer is A. (0). (\(\alpha-3\)\(\beta-3\)=\alpha\beta-3\(\alpha+\beta\)+9). Here (21-30+9=0).
Step 3
Exam Tip
(\(\alpha-3\)\(\beta-3\)=\alpha\beta-3\(\alpha+\beta\)+9) है। यहाँ (21-30+9=0)।
Login to save your score, XP, coins and progress.
Question 258/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2+px+q=0\) के मूल (1) और (p) हैं, तो (q) के लिए कौन-सा संबंध सही है?
If the roots of \(x^2+px+q=0\) are (1) and (p), which relation is correct for (q)?
#quadratic-equations
#roots
#parameter-relation
#hard
A (q=p)
B (q=1+p)
C (q=-p)
D \(q=p^2\)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (q). The given roots are (1) and (p), so \(q=1\cdot p=p\).
Step 2
Why this answer is correct
The correct answer is A. (q=p). The product of roots is (q). The given roots are (1) and (p), so \(q=1\cdot p=p\).
Step 3
Exam Tip
मूलों का गुणनफल (q) होता है। दिए मूल (1) और (p) हैं, इसलिए \(q=1\cdot p=p\)।
Login to save your score, XP, coins and progress.
Question 259/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण \(x^2-4x+1=0\) के मूल \(\alpha,\beta\) हैं। \(\alpha+\beta+\alpha\beta\) का मान क्या है?
The roots of \(x^2-4x+1=0\) are \(\alpha,\beta\). What is \(\alpha+\beta+\alpha\beta\)?
#quadratic-equations
#roots
#expression
#hard
A (5)
B (4)
C (1)
D (3)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (4) and the product is (1). Therefore \(\alpha+\beta+\alpha\beta=5\).
Step 2
Why this answer is correct
The correct answer is A. (5). The sum of roots is (4) and the product is (1). Therefore \(\alpha+\beta+\alpha\beta=5\).
Step 3
Exam Tip
मूलों का योग (4) और गुणनफल (1) है। इसलिए \(\alpha+\beta+\alpha\beta=5\)।
Login to save your score, XP, coins and progress.
Question 260/600
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)?
#quadratic-equations
#root-substitution
#parameter
#hard
A (0)
B \(-\frac{1}{3}\)
C \(\frac{1}{3}\)
D (1)
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)।
Login to save your score, XP, coins and progress.
Question 261/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण ((x-1)2 +(x-2)2 =(x-3)2 ) का मानक रूप कौन-सा है?
What is the standard form of ((x-1)2 +(x-2)2 =(x-3)2 )?
#quadratic-equations
#identity
#standard-form
#hard
A \(x^2+6x-4=0\)
B \(x^2-2x-4=0\)
C \(x^2+2x-4=0\)
D \(x^2-6x+4=0\)
Explanation opens after your attempt
Correct Answer
B. \(x^2-2x-4=0\)
Step 1
Concept
Expanding gives left side \(2x^2-6x+5\) and right side \(x^2-6x+9\). Subtracting gives \(x^2-4=0\).
Step 2
Why this answer is correct
The correct answer is B. \(x^2-2x-4=0\). Expanding gives left side \(2x^2-6x+5\) and right side \(x^2-6x+9\). Subtracting gives \(x^2-4=0\).
Step 3
Exam Tip
विस्तार करने पर बाईं ओर \(2x^2-6x+5\) और दाईं ओर \(x^2-6x+9\) है। घटाने पर \(x^2-4=0\) नहीं बल्कि \(x^2-4=0\) मिलता है।
Login to save your score, XP, coins and progress.
Question 262/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि ((a+1)x-2 +\(a^2-1\)x+2=0) में (a=-1) हो, तो समीकरण किस प्रकार का होगा?
If (a=-1) in ((a+1)x-2 +\(a^2-1\)x+2=0), what type of equation will it be?
#quadratic-equations
#parameter
#degenerate-case
#hard
A द्विघात समीकरण / Quadratic equation
B रैखिक समीकरण / Linear equation
C विरोधाभासी कथन / Contradictory statement
D सदैव सत्य कथन / Always true statement
Explanation opens after your attempt
Correct Answer
C. विरोधाभासी कथन / Contradictory statement
Step 1
Concept
Putting (a=-1) gives \(0x^2+0x+2=0\). This is (2=0), which is a contradictory statement.
Step 2
Why this answer is correct
The correct answer is C. विरोधाभासी कथन / Contradictory statement. Putting (a=-1) gives \(0x^2+0x+2=0\). This is (2=0), which is a contradictory statement.
Step 3
Exam Tip
(a=-1) रखने पर \(0x^2+0x+2=0\) मिलता है। यह (2=0) है, जो विरोधाभासी कथन है।
Login to save your score, XP, coins and progress.
Question 263/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
किस समीकरण में (x=0) एक मूल है और दूसरा मूल धनात्मक है?
In which equation is (x=0) one root and the other root positive?
#quadratic-equations
#zero-root
#root-sign
#hard
A \(x^2-6x=0\)
B \(x^2+6x=0\)
C \(x^2+6=0\)
D \(x^2-6=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x=0\)
Step 1
Concept
(x-2 -6x=x(x-6)), so the roots are (0) and (6). The other root is positive.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x=0\). (x-2 -6x=x(x-6)), so the roots are (0) and (6). The other root is positive.
Step 3
Exam Tip
(x-2 -6x=x(x-6)), इसलिए मूल (0) और (6) हैं। दूसरा मूल धनात्मक है।
Login to save your score, XP, coins and progress.
Question 264/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2+bx+c=0\) के मूल (-2) और (7) हैं, तो (b+c) का मान क्या है?
If the roots of \(x^2+bx+c=0\) are (-2) and (7), what is the value of (b+c)?
#quadratic-equations
#roots
#coefficient-values
#hard
A (-19)
B (19)
C (9)
D (-9)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (5), so (b=-5), and the product is (-14), so (c=-14). Therefore (b+c=-19).
Step 2
Why this answer is correct
The correct answer is A. (-19). The sum of roots is (5), so (b=-5), and the product is (-14), so (c=-14). Therefore (b+c=-19).
Step 3
Exam Tip
मूलों का योग (5) है, इसलिए (b=-5), और गुणनफल (-14) है, इसलिए (c=-14)। अतः (b+c=-19)।
Login to save your score, XP, coins and progress.
Question 265/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण \(x^2-2x-3=0\) के मूलों के घनों का योग क्या है?
What is the sum of cubes of the roots of \(x^2-2x-3=0\)?
#quadratic-equations
#roots
#cubes-sum
#hard
A (26)
B (8)
C (20)
D (-26)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (2) and the product is (-3). (\alpha-3 +\beta-3 =\(\alpha+\beta\)3 -3\alpha\beta\(\alpha+\beta\)=8+18=26).
Step 2
Why this answer is correct
The correct answer is A. (26). The sum of roots is (2) and the product is (-3). (\alpha-3 +\beta-3 =\(\alpha+\beta\)3 -3\alpha\beta\(\alpha+\beta\)=8+18=26).
Step 3
Exam Tip
मूलों का योग (2) और गुणनफल (-3) है। (\alpha-3 +\beta-3 =\(\alpha+\beta\)3 -3\alpha\beta\(\alpha+\beta\)=8+18=26)।
Login to save your score, XP, coins and progress.
Question 266/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(2x^2+kx+18=0\) में मूलों का गुणनफल मूलों के योग का (-3) गुना है, तो (k) क्या होगा?
If in \(2x^2+kx+18=0\), the product of roots is (-3) times the sum of roots, what is (k)?
#quadratic-equations
#roots
#parameter
#hard
A (6)
B (-6)
C (12)
D (-12)
Explanation opens after your attempt
Step 1
Concept
The product is (9) and the sum is \(-\frac{k}{2}\). From (9=-3\left\(-\frac{k}{2}\right\)), we get (k=6).
Step 2
Why this answer is correct
The correct answer is A. (6). The product is (9) and the sum is \(-\frac{k}{2}\). From (9=-3\left\(-\frac{k}{2}\right\)), we get (k=6).
Step 3
Exam Tip
गुणनफल (9) और योग \(-\frac{k}{2}\) है। (9=-3\left\(-\frac{k}{2}\right\)) से (k=6) मिलता है।
Login to save your score, XP, coins and progress.
Question 267/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
किस द्विघात समीकरण के मूलों का योग (0) और गुणनफल (-49) है?
Which quadratic equation has sum of roots (0) and product (-49)?
#quadratic-equations
#sum-product
#forming-equation
#hard
A \(x^2-49=0\)
B \(x^2+49=0\)
C \(x^2+7x-49=0\)
D \(x^2-7x-49=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-49=0\)
Step 1
Concept
\(The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-49) gives (x^2-49=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-49=0). The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-49) gives (x^2-49=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) है। \(योग (0) और गुणनफल (-49) रखने पर (x^2-49=0) मिलता है\)।
Login to save your score, XP, coins and progress.
Question 268/600
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 \(k\neq3\)।
Login to save your score, XP, coins and progress.
Question 269/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
समीकरण \(x^2+px+9=0\) के मूल समान और ऋणात्मक हैं। (p) का मान क्या होगा?
The roots of \(x^2+px+9=0\) are equal and negative. What is the value of (p)?
#quadratic-equations
#equal-negative-roots
#parameter
#hard
A (6)
B (-6)
C (3)
D (-3)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(p^2-36=0\) gives \(p=\pm6\). For equal negative roots, \(-\frac{p}{2}<0\), so (p=6).
Step 2
Why this answer is correct
The correct answer is A. (6). For equal roots, \(p^2-36=0\) gives \(p=\pm6\). For equal negative roots, \(-\frac{p}{2}<0\), so (p=6).
Step 3
Exam Tip
समान मूलों के लिए \(p^2-36=0\) से \(p=\pm6\) मिलता है। ऋणात्मक समान मूल के लिए \(-\frac{p}{2}<0\), इसलिए (p=6)।
Login to save your score, XP, coins and progress.
Question 270/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 29
यदि \(x^2+mx+16=0\) का एक मूल दूसरे का व्युत्क्रम है, तो (m) के बारे में क्या कहा जा सकता है?
If one root of \(x^2+mx+16=0\) is the reciprocal of the other, what can be said about (m)?
#quadratic-equations
#reciprocal-roots
#conceptual
#hard
A ऐसा संभव नहीं है / It is not possible
B (m=16)
C (m=-16)
D (m=0)
Explanation opens after your attempt
Correct Answer
A. ऐसा संभव नहीं है / It is not possible
Step 1
Concept
If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (16), so it is not possible.
Step 2
Why this answer is correct
The correct answer is A. ऐसा संभव नहीं है / It is not possible. If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (16), so it is not possible.
Step 3
Exam Tip
एक मूल दूसरे का व्युत्क्रम हो तो मूलों का गुणनफल (1) होना चाहिए। यहाँ गुणनफल (16) है, इसलिए ऐसा संभव नहीं है।
Login to save your score, XP, coins and progress.
