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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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571 Hard
571 of 914
Analyzing +4 | -1
If α,β\alpha ,\beta are the roots of the equation x2−(5+3log⁡35−5log⁡53)x+3(3(log⁡35)13−5(log⁡53)23−1)=0{x^2} - \left( {5 + {3^{\sqrt {\log _3^5} }} - {5^{\sqrt {\log _5^3} }}} \right)x + 3\left( {{3^{{{\left( {{{\log }_3}5} \right)}^{\frac{1}{3}}}}} - {5^{{{\left( {\log _5^3} \right)}^{\frac{2}{3}}}}} - 1} \right) = 0 then the equation, whose roots are α+1β\alpha+\frac{1}{\beta} and β+1α\beta+\frac{1}{\alpha}
View answer and explanation
Correct option: B
3x2−10x−4=03 x^2-10 x-4=0
Option B (3x2−10x−4=03 x^2-10 x-4=0) is correct because it is derived from substituting the roots into the given equation and simplifying to find the new equation.
572 Easy
572 of 914
Application +4 | -1
The solution set of the equation pqx2−(p+q)2x+(p+q)2=0pq{x^2} - {\left( {p + q} \right)^2}x + {\left( {p + q} \right)^2} = 0 is
View answer and explanation
Correct option: D
{p+qp,  p+qq}\left\{ {\frac{{p + q}}{p},\;\frac{{p + q}}{q}} \right\}
The correct answer is D ({p+qp,  p+qq}\left\{ {\frac{{p + q}}{p},\;\frac{{p + q}}{q}} \right\}) because these are the solutions obtained by solving the given quadratic equation using the quadratic formula or factoring.
573 Easy
573 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of ax2+bx+c=0a x^2+b x+c=0, then those of ax2+2bx+4c=0a x^2+2 b x+4 c=0 are
View answer and explanation
Correct option: B
2α,2β2 \alpha, 2 \beta
The correct answer is B (2α,2β2 \alpha, 2 \beta) because the new roots are twice the original roots, which can be derived by manipulating the quadratic equation.
574 Medium
574 of 914
Analyzing +4 | -1
Let α,β\alpha, \beta be the roots of the equation x2+2x+5=0x^2+2 x+5=0, then the value of 4β3−2β2α3+α\frac{4 \beta^3-2 \beta^2}{\alpha^3+\alpha} is equal to
View answer and explanation
Correct option: B
5
The correct answer is 5 because substituting β\beta into the given equation and simplifying the expression leads to the value 5.
575 Easy
575 of 914
Analyzing +4 | -1
Let α,β\alpha ,\beta be the roots of the equation x2−px+r=0{x^2} - px + r = 0 and α2,  2β\frac{\alpha }{2},\;2\beta be the roots of the equation x2−qx+r=0{x^2} - qx + r = 0. Then the value of rr is
View answer and explanation
Correct option: D
29(2p−q)(2q−p)\frac{2}{9}\left( {2p - q} \right)\left( {2q - p} \right)
Option D (29(2p−q)(2q−p)\frac{2}{9}(2p - q)(2q - p)) is correct because solving the system of equations derived from Vieta's formulas yields this expression for r.
576 Easy
576 of 914
Analyzing +4 | -1
If ax2+bx+c=0a{x^2} + bx + c = 0 and 2x2+3x+4=02{x^2} + 3x + 4 = 0 have a common root where a,  b,  c∈Na,\;b,\;c \in N (set of natural numbers), the least value of a+b+ca + b + c is
View answer and explanation
Correct option: D
9
Option D (9) is correct because the least value of a+b+ca + b + c is 9 when the two quadratic equations have a common root and a,b,ca, b, c are natural numbers.
577 Easy
577 of 914
Analyzing +4 | -1
The number of real roots of the equation (sin⁡2x)(cos⁡2x)=2x+2−x2,\left( {\sin {2^x}} \right)\left( {\cos {2^x}} \right) = \frac{{{2^x} + {2^{ - x}}}}{2}, is
View answer and explanation
Correct option: D
None of these
The correct answer is D. The equation has no solution because the LHS is always less than or equal to 1, while the RHS is always greater than or equal to 2.
578 Medium
578 of 914
Analyzing +4 | -1
If the quadratic equation ax2+bx+6=0a x^2+b x+6=0 does not have real roots and b∈R+b \in R^{+}, then
View answer and explanation
Correct option: A
a>max⁡{b224,b−6}a > \max \left\{\frac{b^2}{24}, b-6\right\}
Option A is correct because it represents the condition where the discriminant is negative, ensuring no real roots, and also satisfies the condition derived from f(-1) > 0.
579 Medium
579 of 914
Application +4 | -1
The number of integer values of xx satisfying (x+1)2>5x−1 and (x+1)2<7x−3(x+1)^2>5 x-1 \text { and }(x+1)^2<7 x-3 simultaneously is
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Correct option: A
1
Option A (1) is correct because only one integer value of x (x=3) satisfies both inequalities simultaneously.
580 Hard
580 of 914
Application +4 | -1
If α,  β\alpha ,\;\beta are the roots of x2−ax+b=0   and  if   αn+βn=Vn,{x^2} - ax + b = 0\;\,\mathrm{and\;if}\,\;{\alpha ^n} + {\beta ^n} = {V_n}, then
View answer and explanation
Correct option: C
Vn+1=aVn−bVn−1{V_{n + 1}} = a{V_n} - b{V_{n - 1}}
Option C (Vn+1=aVn−bVn−1{V_{n + 1}} = a{V_n} - b{V_{n - 1}}) is correct because it is derived by using the given recurrence relation and properties of roots of the quadratic equation.
581 Medium
581 of 914
Application +4 | -1
The product of all the values of xx satisfying the equation (5+26)x2−3+(5−26)x2−3=10(5+2 \sqrt{6})^{x^2-3}+(5-2 \sqrt{6})^{x^2-3}=10 is
View answer and explanation
Correct option: C
8
Option C (8) is correct because the product of all x values satisfying the equation is found by solving for x2 and then multiplying the roots.
582 Easy
582 of 914
Analyzing +4 | -1
Both the roots of the given equation (x−a)(x−b)+(x−b)(x−c)+(x−c)(x−a)=0\left( {x - a} \right)\left( {x - b} \right) + \left( {x - b} \right)\left( {x - c} \right) + \left( {x - c} \right)\left( {x - a} \right) = 0 are always
View answer and explanation
Correct option: C
Real
Option C (Real) is correct because the discriminant of the quadratic equation is always non-negative, ensuring that the roots are real.
583 Easy
583 of 914
Application +4 | -1
The values of &#39; aa &#39; for which one root of the equation x2−(a+1)x+a2+a−8=0x^{2}-(a+1) x+a^{2}+a-8=0 exceeds 2 and the other is lesser than 2 , are given by :
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Correct option: C
&nbsp;-2
Option C (&nbsp;-2) is correct because the condition that one root exceeds 2 and the other is less than 2 implies f(2)<0f(2) < 0, which leads to the solution a<−2a < -2.
584 Easy
584 of 914
Analyzing +4 | -1
The solution set of the inequation ∣2x−1x−1∣>2,\left| {\frac{{2x - 1}}{{x - 1}}} \right| > 2, is
View answer and explanation
Correct option: A
(3/4  ,1)∪(1,∞)\left( {3/4\;,1} \right) \cup \left( {1,\infty } \right)
Option A (3/4  ,1)∪(1,∞)\left( {3/4\;,1} \right) \cup \left( {1,\infty } \right) is the correct solution set obtained by solving the inequality by considering both positive and negative cases of the absolute value.
585 Medium
585 of 914
Application +4 | -1
Given that tan⁡A\tan A and tan⁡B\tan B are the roots of x2−px+q=0{x^2} - px + q = 0, then the value of sin⁡2(A+B){\sin ^2}\left( {A + B} \right) is
View answer and explanation
Correct option: A
p2p2(1−q)2\frac{{{p^2}}}{{{p^2}{{\left( {1 - q} \right)}^2}}}
Option D (p2p2+(1−q)2)\frac{{{p^2}}}{{{p^2} + {{\left( {1 - q} \right)}^2}}}) is correct because using the relationship between the roots and coefficients of the quadratic equation, we can find tan⁡(A+B)\tan(A+B) and then use trigonometric identities to find sin⁡2(A+B)\sin^2(A+B).
586 Medium
586 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0 then, (1+α+α2)  (1+β+β2)\left( {1 + \alpha + {\alpha ^2}} \right)\;\left( {1 + \beta + {\beta ^2}} \right) is equal to
View answer and explanation
Correct option: B
Positive
The correct answer is A (x=4,  9x = 4,\;9) because these are the only values that satisfy the given absolute value equation.
587 Easy
587 of 914
Application +4 | -1
(log⁡5x)2+(log⁡5x)<2,{\left( {{{\log }_5}x} \right)^2} + \left( {{{\log }_5}x} \right) < 2, then xx belongs to the interval
View answer and explanation
Correct option: A
(1/25  ,5)\left( {1/25\;,5} \right)
Option A ((1/25  ,5)\left( {1/25\;,5} \right)) is correct because it represents the interval to which x belongs after solving the logarithmic inequality.
588 Medium
588 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of the equation ax2+bx+c=0a x^2+b x+c=0, then the equation whose roots are 1α+β,1α+1β\frac{1}{\alpha+\beta}, \frac{1}{\alpha}+\frac{1}{\beta} is
View answer and explanation
Correct option: B
bcx2+(b2+ac)x+ab=0b c x^2+\left(b^2+a c\right) x+a b=0
Option B (bcx2+(b2+ac)x+ab=0b c x^2+\left(b^2+a c\right) x+a b=0) is correct because it is derived from the sum and product of the new roots formed from the original roots α and β.
589 Easy
589 of 914
Analyzing +4 | -1
If a,  ba,\;b and cc are in geometric progression and the roots of the equation ax2+2bx+c=0a{x^2} + 2bx + c = 0 are α\alpha and β\beta and those of cx2+2bx+a=0c{x^2} + 2bx + a = 0 are γ\gamma and δ\delta, then
View answer and explanation
Correct option: C
aα=aβ=cγ=cδa\alpha = a\beta = c\gamma = c\delta
Option D (aα=aβ=cγ=cδa\alpha = a\beta = c\gamma = c\delta) is correct because the roots of both quadratic equations are equal and related by the common ratio of the geometric progression.
590 Easy
590 of 914
Application +4 | -1
The sum of all real roots of the equation ∣x−2∣2+∣x−2∣−2=0|x-2|^2+|x-2|-2=0 is
View answer and explanation
Correct option: B
4
Option B (4) is correct because by substituting t=∣x−2∣t = |x-2|, we find the real roots x=1x=1 and x=3x=3, and their sum is 4.
591 Medium
591 of 914
Application +4 | -1
The value of aa for which the sum of the squares of the roots of the equation x2−(a−2)x−a−1=0{x^2} - \left( {a - 2} \right)x - a - 1 = 0 assumes the least value, is
View answer and explanation
Correct option: B
1
The correct answer is B (1) because the minimum value of the sum of the squares of the roots occurs when a = 1.
592 Easy
592 of 914
Application +4 | -1
If ecos⁡x−e−cos⁡x=4,{e^{\cos x}} - {e^{ - \cos x}} = 4, then the value of cos⁡x\cos x is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the derived value of cos⁡x\cos x is greater than 1, which is not possible for the cosine function.
593 Medium
593 of 914
Application +4 | -1
If ai>0a_i > 0 for i=1,2,…,ni=1,2, \ldots, n and a1a2…an=1a_1 a_2 \ldots a_n=1, then minimum value of (1+a1)(1+a2)…(1+an)\left(1+a_1\right)\left(1+a_2\right) \ldots\left(1+a_n\right) is
View answer and explanation
Correct option: B
2n2^n
The correct answer is B (2n2^n) because it is found by applying the AM-GM inequality to each (1+ai)(1+a_i) term and then multiplying the resulting inequalities.
594 Easy
594 of 914
Application +4 | -1
If α,β\alpha, \beta are roots of 9x2−11x+1=09 x^2-11 x+1=0 then value of 1(9α−11)2+(11β−19)\frac{1}{(9 \alpha-11)^2}+\left(\frac{11 \beta-1}{9}\right) is-
View answer and explanation
Correct option: D
10381\frac{103}{81}
Option D (10381\frac{103}{81}) is correct because by using Vieta's formulas and the given quadratic equation, the expression simplifies to α2+β2=10381\alpha^2 + \beta^2 = \frac{103}{81}.
595 Medium
595 of 914
Application +4 | -1
If aa and bb are the non-zero distinct roots of x2+ax+b=0,{x^2} + ax + b = 0, then the least value of x2+ax+b{x^2} + ax + b is
View answer and explanation
Correct option: C
−94- \frac{9}{4}
Option C (-9/4) is correct because it is the minimum value of the quadratic x2+ax+b{x^2} + ax + b given the conditions on a and b.
596 Easy
596 of 914
Application +4 | -1
If α+β=4\alpha + \beta = 4 and α3+β3=44{\alpha ^3} + {\beta ^3} = 44, then α,  β\alpha ,\;\beta are the roots of the equation
View answer and explanation
Correct option: D
3x2−12x+5=03{x^2} - 12x + 5 = 0
Option D (3x² - 12x + 5 = 0) is correct because using the given conditions to find the sum and product of the roots leads to this quadratic equation.
597 Medium
597 of 914
Application +4 | -1
Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots (4,3)(4,3). Rahul made a mistake in writing down coefficient of xx to get roots (3,2)(3,2). The correct roots of equation are
View answer and explanation
Correct option: B
6,1
Option B (6,1) is correct because Sachin's mistake affects the product of roots, and Rahul's mistake affects the sum of roots, allowing us to reconstruct the correct quadratic equation.
598 Easy
598 of 914
Application +4 | -1
If a,c≠0a,c \ne 0 and α,β\alpha ,\beta are the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0, then the quadratic equation with 1/α1/\alpha and 1/β1/\beta as its root is
View answer and explanation
Correct option: B
cx2+bx+a=0c{x^2} + bx + a = 0
Option B (cx2+bx+a=0c{x^2} + bx + a = 0) is correct because it represents the quadratic equation with roots that are reciprocals of the original roots.
599 Easy
599 of 914
Application +4 | -1
x+4x−3<2\frac{x+4}{x-3} < 2 is satisfied when xx satisfies
View answer and explanation
Correct option: A
(−∞,3)∪(10,∞)(-\infty, 3) \cup(10, \infty)
Option A ((−∞,3)∪(10,∞))((-\infty, 3) \cup (10, \infty)) is correct because it represents the set of xx values that satisfy the given inequality.
600 Easy
600 of 914
Application +4 | -1
If one root of the quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 is equal to nth power of the other root, then the value of (acn)1/(n+1)+(anc)1/(n+1){\left( {a{c^n}} \right)^{1/\left( {n + 1} \right)}} + {\left( {{a^n}c} \right)^{1/\left( {n + 1} \right)}} is equal to
View answer and explanation
Correct option: B
−b- b
Option B (−b-b) is correct because it is the result of eliminating α from the sum and product of the roots, and then simplifying the expression.
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