Question 211/600
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)?
#quadratic-equations
#root-substitution
#parameter
#hard
A \( \frac{1}{2} \)
B \( -\frac{1}{2} \)
C (1)
D (0)
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 212/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण ((x+2)2 +(x-5)2 =(x+1)2 ) का मानक रूप कौन-सा है?
What is the standard form of ((x+2)2 +(x-5)2 =(x+1)2 )?
#quadratic-equations
#identity
#standard-form
#hard
A \(x^2-8x+28=0\)
B \(x^2+8x+28=0\)
C \(x^2-6x+28=0\)
D \(x^2-8x-28=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-8x+28=0\)
Step 1
Concept
Expanding gives left side \(2x^2-6x+29\) and right side \(x^2+2x+1\). Subtracting gives \(x^2-8x+28=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-8x+28=0\). Expanding gives left side \(2x^2-6x+29\) and right side \(x^2+2x+1\). Subtracting gives \(x^2-8x+28=0\).
Step 3
Exam Tip
विस्तार करने पर बाईं ओर \(2x^2-6x+29\) और दाईं ओर \(x^2+2x+1\) है। घटाने पर \(x^2-8x+28=0\) मिलता है।
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Question 213/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (a=2) हो, तो ((a-2)x-2 +\(a^2-4\)x+5=0) किस प्रकार का कथन बनेगा?
If (a=2), what type of statement will ((a-2)x-2 +\(a^2-4\)x+5=0) become?
#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=2) gives \(0x^2+0x+5=0\). This is (5=0), which is a contradictory statement.
Step 2
Why this answer is correct
The correct answer is C. विरोधाभासी कथन / Contradictory statement. Putting (a=2) gives \(0x^2+0x+5=0\). This is (5=0), which is a contradictory statement.
Step 3
Exam Tip
(a=2) रखने पर \(0x^2+0x+5=0\) मिलता है। यह (5=0) है, जो विरोधाभासी कथन है।
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Question 214/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
किस समीकरण में (x=0) एक मूल है और दूसरा मूल ऋणात्मक है?
In which equation is (x=0) one root and the other root negative?
#quadratic-equations
#zero-root
#root-sign
#hard
A \(x^2+7x=0\)
B \(x^2-7x=0\)
C \(x^2+7=0\)
D \(x^2-7=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+7x=0\)
Step 1
Concept
(x-2 +7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+7x=0\). (x-2 +7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.
Step 3
Exam Tip
(x-2 +7x=x(x+7)), इसलिए मूल (0) और (-7) हैं। दूसरा मूल ऋणात्मक है।
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Question 215/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2+bx+c=0\) के मूल (-3) और (8) हैं, तो (b+c) का मान क्या है?
If the roots of \(x^2+bx+c=0\) are (-3) and (8), what is the value of (b+c)?
#quadratic-equations
#roots
#coefficient-values
#hard
A (-29)
B (29)
C (19)
D (-19)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (5), so (b=-5), and the product is (-24), so (c=-24). Therefore (b+c=-29).
Step 2
Why this answer is correct
The correct answer is A. (-29). The sum of roots is (5), so (b=-5), and the product is (-24), so (c=-24). Therefore (b+c=-29).
Step 3
Exam Tip
मूलों का योग (5) है, इसलिए (b=-5), और गुणनफल (-24) है, इसलिए (c=-24)। अतः (b+c=-29)।
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Question 216/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(x^2-3x-10=0\) के मूलों के घनों का योग क्या है?
What is the sum of cubes of the roots of \(x^2-3x-10=0\)?
#quadratic-equations
#roots
#cubes-sum
#hard
A (117)
B (27)
C (90)
D (-117)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (3) and the product is (-10). (\alpha-3 +\beta-3 =\(\alpha+\beta\)3 -3\alpha\beta\(\alpha+\beta\)=27+90=117).
Step 2
Why this answer is correct
The correct answer is A. (117). The sum of roots is (3) and the product is (-10). (\alpha-3 +\beta-3 =\(\alpha+\beta\)3 -3\alpha\beta\(\alpha+\beta\)=27+90=117).
Step 3
Exam Tip
मूलों का योग (3) और गुणनफल (-10) है। (\alpha-3 +\beta-3 =\(\alpha+\beta\)3 -3\alpha\beta\(\alpha+\beta\)=27+90=117)।
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Question 217/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(3x^2+kx+12=0\) में मूलों का गुणनफल मूलों के योग का (-2) गुना है, तो (k) क्या होगा?
If in \(3x^2+kx+12=0\), the product of roots is (-2) 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 (4) and the sum is \(-\frac{k}{3}\). From (4=-2\left\(-\frac{k}{3}\right\)), we get (k=6).
Step 2
Why this answer is correct
The correct answer is A. (6). The product is (4) and the sum is \(-\frac{k}{3}\). From (4=-2\left\(-\frac{k}{3}\right\)), we get (k=6).
Step 3
Exam Tip
गुणनफल (4) और योग \(-\frac{k}{3}\) है। (4=-2\left\(-\frac{k}{3}\right\)) से (k=6) मिलता है।
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Question 218/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
किस द्विघात समीकरण के मूलों का योग (0) और गुणनफल (-81) है?
Which quadratic equation has sum of roots (0) and product (-81)?
#quadratic-equations
#sum-product
#forming-equation
#hard
A \(x^2-81=0\)
B \(x^2+81=0\)
C \(x^2+9x-81=0\)
D \(x^2-9x-81=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-81=0\)
Step 1
Concept
\(The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-81) gives (x^2-81=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-81=0). The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-81) gives (x^2-81=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) है। \(योग (0) और गुणनफल (-81) रखने पर (x^2-81=0) मिलता है\)।
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Question 219/600
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 220/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(x^2+px+16=0\) के मूल समान और धनात्मक हैं। (p) का मान क्या होगा?
The roots of \(x^2+px+16=0\) are equal and positive. What is the value of (p)?
#quadratic-equations
#equal-positive-roots
#parameter
#hard
A (-8)
B (8)
C (4)
D (-4)
Explanation opens after your attempt
Step 1
Concept
For equal roots, \(p^2-64=0\) gives \(p=\pm8\). For equal positive roots, \(-\frac{p}{2}>0\), so (p=-8).
Step 2
Why this answer is correct
The correct answer is A. (-8). For equal roots, \(p^2-64=0\) gives \(p=\pm8\). For equal positive roots, \(-\frac{p}{2}>0\), so (p=-8).
Step 3
Exam Tip
समान मूलों के लिए \(p^2-64=0\) से \(p=\pm8\) मिलता है। धनात्मक समान मूल के लिए \(-\frac{p}{2}>0\), इसलिए (p=-8)।
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Question 221/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2+mx+25=0\) का एक मूल दूसरे का व्युत्क्रम है, तो (m) के बारे में क्या कहा जा सकता है?
If one root of \(x^2+mx+25=0\) 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=25)
C (m=-25)
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 (25), 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 (25), so it is not possible.
Step 3
Exam Tip
एक मूल दूसरे का व्युत्क्रम हो तो मूलों का गुणनफल (1) होना चाहिए। यहाँ गुणनफल (25) है, इसलिए ऐसा संभव नहीं है।
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Question 222/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(9x^2+12kx+4k^2=0\) के मूलों की प्रकृति क्या है?
What is the nature of roots of \(9x^2+12kx+4k^2=0\)?
#quadratic-equations
#perfect-square
#nature-of-roots
#hard
A दो समान वास्तविक मूल / Two equal real roots
B दो भिन्न वास्तविक मूल / Two distinct real roots
C कोई वास्तविक मूल नहीं / No real roots
D निर्धारित नहीं हो सकता / Cannot be determined
Explanation opens after your attempt
Correct Answer
A. दो समान वास्तविक मूल / Two equal real roots
Step 1
Concept
It can be written as ((3x+2k)2 =0). Therefore both roots are equal and real.
Step 2
Why this answer is correct
The correct answer is A. दो समान वास्तविक मूल / Two equal real roots. It can be written as ((3x+2k)2 =0). Therefore both roots are equal and real.
Step 3
Exam Tip
यह ((3x+2k)2 =0) के रूप में लिखा जा सकता है। इसलिए दोनों मूल समान वास्तविक होते हैं।
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Question 223/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि ((x+a)2 =x-2 -12x+36) है, तो (a) का मान क्या होगा?
If ((x+a)2 =x-2 -12x+36), what is the value of (a)?
#quadratic-equations
#identity
#coefficient-comparison
#hard
A (-6)
B (6)
C (-12)
D (36)
Explanation opens after your attempt
Step 1
Concept
((x+a)2 =x-2 +2ax+a-2 ). From (2a=-12), we get (a=-6).
Step 2
Why this answer is correct
The correct answer is A. (-6). ((x+a)2 =x-2 +2ax+a-2 ). From (2a=-12), we get (a=-6).
Step 3
Exam Tip
((x+a)2 =x-2 +2ax+a-2 ) होता है। (2a=-12) से (a=-6) मिलता है।
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Question 224/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
किस समीकरण में मूल बराबर और ऋणात्मक होंगे?
Which equation will have equal and negative roots?
#quadratic-equations
#equal-negative-roots
#perfect-square
#hard
A \(x^2+16x+64=0\)
B \(x^2-16x+64=0\)
C \(x^2-64=0\)
D \(x^2+64=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+16x+64=0\)
Step 1
Concept
(x-2 +16x+64=(x+8)2 ), so both roots are (-8). Both roots are equal and negative.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+16x+64=0\). (x-2 +16x+64=(x+8)2 ), so both roots are (-8). Both roots are equal and negative.
Step 3
Exam Tip
(x-2 +16x+64=(x+8)2 ), इसलिए दोनों मूल (-8) हैं। दोनों मूल बराबर और ऋणात्मक हैं।
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Question 225/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2-7x+12=0\) के मूल \(\alpha,\beta\) हैं, तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) क्या है?
If \(\alpha,\beta\) are roots of \(x^2-7x+12=0\), what is \(\frac{1}{\alpha}+\frac{1}{\beta}\)?
#quadratic-equations
#reciprocal-roots
#hard
A \( \frac{7}{12} \)
B \( \frac{12}{7} \)
C (7)
D (12)
Explanation opens after your attempt
Correct Answer
A. \( \frac{7}{12} \)
Step 1
Concept
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the value is \(\frac{7}{12}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{7}{12} \). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the value is \(\frac{7}{12}\).
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ मान \(\frac{7}{12}\) है।
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Question 226/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
दो संख्याओं का योग (13) और गुणनफल (40) है। वे किस द्विघात समीकरण के मूल हो सकते हैं?
Two numbers have sum (13) and product (40). They can be roots of which quadratic equation?
#quadratic-equations
#forming-equation
#sum-product
#hard
A \(x^2-13x+40=0\)
B \(x^2+13x+40=0\)
C \(x^2-40x+13=0\)
D \(x^2+40x-13=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-13x+40=0\)
Step 1
Concept
If the sum of roots is (13) and product is (40), the equation is \(x^2-13x+40=0\). Remember the monic form formula.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-13x+40=0\). If the sum of roots is (13) and product is (40), the equation is \(x^2-13x+40=0\). Remember the monic form formula.
Step 3
Exam Tip
यदि मूलों का योग (13) और गुणनफल (40) है, तो समीकरण \(x^2-13x+40=0\) होगा। मोनिक रूप का सूत्र याद रखें।
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Question 227/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
एक आयत की लंबाई (x+7) और चौड़ाई (x-3) है। क्षेत्रफल (90) हो तो सही समीकरण कौन-सा है?
A rectangle has length (x+7) and breadth (x-3). If its area is (90), which equation is correct?
#quadratic-equations
#word-problem
#area
#hard
A \(x^2+4x-111=0\)
B \(x^2+4x-90=0\)
C \(x^2-4x-111=0\)
D \(x^2+10x-111=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+4x-111=0\)
Step 1
Concept
The area is ((x+7)(x-3)=90). Expanding gives \(x^2+4x-21=90\) and \(x^2+4x-111=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+4x-111=0\). The area is ((x+7)(x-3)=90). Expanding gives \(x^2+4x-21=90\) and \(x^2+4x-111=0\).
Step 3
Exam Tip
क्षेत्रफल ((x+7)(x-3)=90) होगा। विस्तार से \(x^2+4x-21=90\) और \(x^2+4x-111=0\) मिलता है।
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Question 228/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
किस विकल्प में मूलों का योग ऋणात्मक और गुणनफल धनात्मक होगा?
In which option will the sum of roots be negative and the product of roots be positive?
#quadratic-equations
#sum-product
#signs
#hard
A \(x^2+7x+12=0\)
B \(x^2-7x+12=0\)
C \(x^2+7x-12=0\)
D \(x^2-7x-12=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+7x+12=0\)
Step 1
Concept
In the first option, the sum is \(-\frac{b}{a}=-7\) and the product is \(\frac{c}{a}=12\). So the sum is negative and the product is positive.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+7x+12=0\). In the first option, the sum is \(-\frac{b}{a}=-7\) and the product is \(\frac{c}{a}=12\). So the sum is negative and the product is positive.
Step 3
Exam Tip
पहले विकल्प में योग \(-\frac{b}{a}=-7\) और गुणनफल \(\frac{c}{a}=12\) है। इसलिए योग ऋणात्मक और गुणनफल धनात्मक है।
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Question 229/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि समीकरण \(x^2+bx+49=0\) के दोनों मूल समान हैं और उनका मान (-7) है, तो (b) क्या है?
If both roots of \(x^2+bx+49=0\) are equal and their value is (-7), what is (b)?
#quadratic-equations
#equal-roots
#coefficient
#hard
A (14)
B (-14)
C (7)
D (-7)
Explanation opens after your attempt
Step 1
Concept
Both roots are (-7), so their sum is (-14). Here the sum of roots is (-b), so (b=14).
Step 2
Why this answer is correct
The correct answer is A. (14). Both roots are (-7), so their sum is (-14). Here the sum of roots is (-b), so (b=14).
Step 3
Exam Tip
दोनों मूल (-7) हैं, इसलिए योग (-14) है। यहाँ मूलों का योग (-b) है, अतः (b=14)।
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Question 230/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2-14x+c=0\) पूर्ण वर्ग द्विघात समीकरण है, तो (c) का मान क्या होगा?
If \(x^2-14x+c=0\) is a perfect square quadratic equation, what is the value of (c)?
#quadratic-equations
#perfect-square
#parameter
#hard
A (49)
B (14)
C (28)
D (196)
Explanation opens after your attempt
Step 1
Concept
For a perfect square, (x-2 -14x+c=(x-7)2 ) is needed. Hence (c=49).
Step 2
Why this answer is correct
The correct answer is A. (49). For a perfect square, (x-2 -14x+c=(x-7)2 ) is needed. Hence (c=49).
Step 3
Exam Tip
पूर्ण वर्ग के लिए (x-2 -14x+c=(x-7)2 ) होना चाहिए। इसलिए (c=49) होगा।
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Question 231/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(2x^2-9x+10=0\) में मूलों का अंतर क्या है?
What is the difference between the roots of \(2x^2-9x+10=0\)?
#quadratic-equations
#roots
#difference
#hard
A \( \frac{1}{2} \)
B \( \frac{3}{2} \)
C (2)
D \( \frac{5}{2} \)
Explanation opens after your attempt
Correct Answer
A. \( \frac{1}{2} \)
Step 1
Concept
The roots are (2) and \(\frac{5}{2}\). Their difference is \(\frac{5}{2}-2=\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{1}{2} \). The roots are (2) and \(\frac{5}{2}\). Their difference is \(\frac{5}{2}-2=\frac{1}{2}\).
Step 3
Exam Tip
मूल (2) और \(\frac{5}{2}\) हैं। उनका अंतर \(\frac{5}{2}-2=\frac{1}{2}\) है।
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Question 232/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि (x-2 -(m+4)x+4m=0) का एक मूल (4) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+4)x+4m=0) is (4), what is the other root?
#quadratic-equations
#one-root
#roots
#hard
A (m)
B (4m)
C (m+4)
D (m-4 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (4m) and one root is (4). Hence the other root is \(\frac{4m}{4}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (4m) and one root is (4). Hence the other root is \(\frac{4m}{4}=m\).
Step 3
Exam Tip
गुणनफल (4m) है और एक मूल (4) है। इसलिए दूसरा मूल \(\frac{4m}{4}=m\) है।
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Question 233/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(x^2-6x+11=0\) के लिए कौन-सा कथन सही है?
Which statement is correct for \(x^2-6x+11=0\)?
#quadratic-equations
#nature-of-roots
#hard
A वास्तविक मूल नहीं हैं / It has no real roots
B दो समान वास्तविक मूल हैं / It has two equal real roots
C दो भिन्न वास्तविक मूल हैं / It has two distinct real roots
D एकमात्र वास्तविक मूल है / It has only one real root
Explanation opens after your attempt
Correct Answer
A. वास्तविक मूल नहीं हैं / It has no real roots
Step 1
Concept
Here (D=(-6)2 -4\cdot1\cdot11=-8<0). Therefore it has no real roots.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं हैं / It has no real roots. Here (D=(-6)2 -4\cdot1\cdot11=-8<0). Therefore it has no real roots.
Step 3
Exam Tip
यहाँ (D=(-6)2 -4\cdot1\cdot11=-8<0) है। इसलिए वास्तविक मूल नहीं होंगे।
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Question 234/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि किसी मोनिक द्विघात समीकरण के मूल (r) और (2r) हैं तथा उनका योग (9) है, तो उस समीकरण में स्थिर पद क्या होगा?
If the roots of a monic quadratic equation are (r) and (2r), and their sum is (9), what will be the constant term?
#quadratic-equations
#roots
#monic-equation
#hard
A (18)
B (9)
C (27)
D (36)
Explanation opens after your attempt
Step 1
Concept
From (r+2r=9), we get (r=3), so the roots are (3) and (6). The constant term will be the product of roots (18).
Step 2
Why this answer is correct
The correct answer is A. (18). From (r+2r=9), we get (r=3), so the roots are (3) and (6). The constant term will be the product of roots (18).
Step 3
Exam Tip
(r+2r=9) से (r=3) मिलता है, इसलिए मूल (3) और (6) हैं। स्थिर पद मूलों का गुणनफल (18) होगा।
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Question 235/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण ((x+2)(x+5)+(x-1)(x+4)=60) का मानक रूप कौन-सा है?
What is the standard form of ((x+2)(x+5)+(x-1)(x+4)=60)?
#quadratic-equations
#expansion
#standard-form
#hard
A \(2x^2+10x-54=0\)
B \(2x^2+10x+6=0\)
C \(x^2+5x-54=0\)
D \(2x^2+5x-60=0\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2+10x-54=0\)
Step 1
Concept
The left side is \(x^2+7x+10+x^2+3x-4=2x^2+10x+6\). Subtracting (60) gives \(2x^2+10x-54=0\).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2+10x-54=0\). The left side is \(x^2+7x+10+x^2+3x-4=2x^2+10x+6\). Subtracting (60) gives \(2x^2+10x-54=0\).
Step 3
Exam Tip
बाईं ओर \(x^2+7x+10+x^2+3x-4=2x^2+10x+6\) है। (60) घटाने पर \(2x^2+10x-54=0\) मिलता है।
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Question 236/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2+7x+10=0\) के मूल \(\alpha\) और \(\beta\) हैं, तो (\(\alpha+2\)\(\beta+2\)) का मान क्या है?
If \(\alpha\) and \(\beta\) are roots of \(x^2+7x+10=0\), what is (\(\alpha+2\)\(\beta+2\))?
#quadratic-equations
#roots
#expression-value
#hard
A (0)
B (10)
C (4)
D (28)
Explanation opens after your attempt
Step 1
Concept
(\(\alpha+2\)\(\beta+2\)=\alpha\beta+2\(\alpha+\beta\)+4). Here (10+2(-7)+4=0).
Step 2
Why this answer is correct
The correct answer is A. (0). (\(\alpha+2\)\(\beta+2\)=\alpha\beta+2\(\alpha+\beta\)+4). Here (10+2(-7)+4=0).
Step 3
Exam Tip
(\(\alpha+2\)\(\beta+2\)=\alpha\beta+2\(\alpha+\beta\)+4) होता है। यहाँ (10+2(-7)+4=0)।
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Question 237/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(3x^2+8x+4=0\) में मूलों के वर्गों का योग क्या है?
What is the sum of squares of the roots of \(3x^2+8x+4=0\)?
#quadratic-equations
#roots
#squares-sum
#hard
A \( \frac{40}{9} \)
B \( \frac{64}{9} \)
C \( \frac{16}{9} \)
D \( \frac{28}{9} \)
Explanation opens after your attempt
Correct Answer
A. \( \frac{40}{9} \)
Step 1
Concept
If the roots are \(\alpha,\beta\), then (\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta). Here (\left\(-\frac{8}{3}\right\)2 -2\cdot\frac{4}{3}=\frac{40}{9}).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{40}{9} \). If the roots are \(\alpha,\beta\), then (\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta). Here (\left\(-\frac{8}{3}\right\)2 -2\cdot\frac{4}{3}=\frac{40}{9}).
Step 3
Exam Tip
यदि मूल \(\alpha,\beta\) हैं, तो (\alpha-2 +\beta-2 =\(\alpha+\beta\)2 -2\alpha\beta)। यहाँ (\left\(-\frac{8}{3}\right\)2 -2\cdot\frac{4}{3}=\frac{40}{9})।
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Question 238/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
यदि \(x^2+ax+18=0\) का एक मूल (2) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+18=0\) is (2), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#hard
A दूसरा मूल (9), (a=-11) / other root (9), (a=-11)
B दूसरा मूल (-9), (a=7) / other root (-9), (a=7)
C दूसरा मूल (9), (a=11) / other root (9), (a=11)
D दूसरा मूल (-2), (a=0) / other root (-2), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (9), (a=-11) / other root (9), (a=-11)
Step 1
Concept
The product of roots is (18), so the other root is (9). The sum is (11), and (-a=11), so (a=-11).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (9), (a=-11) / other root (9), (a=-11). The product of roots is (18), so the other root is (9). The sum is (11), and (-a=11), so (a=-11).
Step 3
Exam Tip
मूलों का गुणनफल (18) है, इसलिए दूसरा मूल (9) होगा। योग (11) है और (-a=11), इसलिए (a=-11)।
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Question 239/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
किस समीकरण का विवेचक (-11) है?
Which equation has discriminant (-11)?
#quadratic-equations
#discriminant
#identify
#hard
A \(x^2+x+3=0\)
B \(x^2+3x+1=0\)
C \(2x^2+x+2=0\)
D \(x^2-4x+10=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+x+3=0\)
Step 1
Concept
For \(x^2+x+3=0\), \(D=1^2-4\cdot1\cdot3=-11\). Subtract the full (4ac) while finding the discriminant.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+x+3=0\). For \(x^2+x+3=0\), \(D=1^2-4\cdot1\cdot3=-11\). Subtract the full (4ac) while finding the discriminant.
Step 3
Exam Tip
\(x^2+x+3=0\) के लिए \(D=1^2-4\cdot1\cdot3=-11\) है। विवेचक निकालते समय (4ac) पूरा घटाएं।
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Question 240/600
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
समीकरण \(x^2-8x+k=0\) के मूल वास्तविक और भिन्न हों, तो (k) पर सही शर्त क्या है?
If the roots of \(x^2-8x+k=0\) are real and distinct, what is the correct condition on (k)?
#quadratic-equations
#distinct-real-roots
#parameter
#hard
A (k<16)
B (k=16)
C (k>16)
D \(k\leq16\)
Explanation opens after your attempt
Step 1
Concept
For real and distinct roots, (D>0) is needed. Here (64-4k>0), so (k<16).
Step 2
Why this answer is correct
The correct answer is A. (k<16). For real and distinct roots, (D>0) is needed. Here (64-4k>0), so (k<16).
Step 3
Exam Tip
भिन्न वास्तविक मूलों के लिए (D>0) चाहिए। यहाँ (64-4k>0), इसलिए (k<16)।
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