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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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Practice format and chapter quiz

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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Showing 1-30 of 914 questions
1 Easy
1 of 914
Application +4 | -1
The solution of the equation log⁡7log⁡5(x+5+x)=0\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0 is
View answer and explanation
Correct option: A
4
Option A (4) is correct because solving the logarithmic equation leads to x = 4.
2 Easy
2 of 914
Analyzing +4 | -1
If α,β\alpha ,\beta are the roots of x2−3x+a=0,a∈R{x^2} - 3x + a = 0,a \in R and α<1<β,\alpha < 1 < \beta , then
View answer and explanation
Correct option: A
a∈(−∞,2)a \in \left( { - \infty ,2} \right)
Option A (a∈(−∞,2)a \in ( - \infty ,2)) is correct because for α<1<β\alpha < 1 < \beta, f(1)<0f(1) < 0 must hold, leading to the condition a<2a < 2.
3 Easy
3 of 914
Analyzing +4 | -1
If α,  β\alpha ,\;\beta be the roots of the equation x2−px+q=0{x^2} - px + q = 0 and α1,β1{\alpha _1},{\beta _1} be the roots of the equation x2−qx+p=0,{x^2} - qx + p = 0, then the equation whose roots are 1α1β+1αβ1and  1αα1+1ββ1,\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}}, is
View answer and explanation
Correct option: B
p2q2x2−p2q2x+p3+q3−4pq=0{p^2}{q^2}{x^2} - {p^2}{q^2}x + {p^3} + {q^3} - 4pq = 0
Option B p2q2x2−p2q2x+p3+q3−4pq=0{p^2}{q^2}{x^2} - {p^2}{q^2}x + {p^3} + {q^3} - 4pq = 0 is correct because it is derived by finding the sum and product of the new roots and then constructing the quadratic equation.
4 Medium
4 of 914
Application +4 | -1
If x2+4ax+2>0{x^2} + 4ax + 2 > 0 for all values of xx, then aa lies in the interval
View answer and explanation
Correct option: D
(−12,  12)\left( { - \frac{1}{{\sqrt 2 }},\;\frac{1}{{\sqrt 2 }}} \right)
Option D ((−12,  12)\left( { - \frac{1}{{\sqrt 2 }},\;\frac{1}{{\sqrt 2 }}} \right)) is correct because for the quadratic to be always positive, the discriminant must be negative, leading to the given interval for aa.
5 Medium
5 of 914
Analyzing +4 | -1
If f(x)=ax2+bx+c  f\left( x \right) = a{x^2} + bx + c\; and   f(−1)<1,f(1)>−1,f(3)<−4  \;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\; and   a≠0,\;a \ne 0, then
View answer and explanation
Correct option: B
a<0a < 0
Option B (a<0a < 0) is correct because the given conditions on f(-1), f(1), and f(3) lead to a negative value for 'a' after solving the inequalities.
6 Hard
6 of 914
Analyzing +4 | -1
Root(s) of the equation 9x2−18∣x∣+5=09{x^2} - 18\left| x \right| + 5 = 0 belonging to the domain of definition of the function f(x)=log⁡(x2−x−2)f\left( x \right) = \log \left( {{x^2} - x - 2} \right) is (are)
View answer and explanation
Correct option: C
−53\frac{{ - 5}}{3}
Option C (−53\frac{{ - 5}}{3}) is correct because it is the only root of the given equation that also satisfies the domain of the logarithmic function.
7 Easy
7 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, αβ=3\alpha \beta = 3 and a,  b,  ca,\;b,\;c are in AP, then α+β\alpha + \beta is equal to
View answer and explanation
Correct option: D
−2- 2
Option D (−2-2) is correct because by using the given condition αβ=3\alpha\beta = 3 and the fact that a, b, c are in AP, we can find the value of α+β\alpha + \beta.
8 Medium
8 of 914
Analyzing +4 | -1
The solution set of the inequation 1∣x∣−3<12,\frac{1}{{\left| x \right| - 3}} < \frac{1}{2}, is
View answer and explanation
Correct option: C
(−∞,  −5)∪(−3,  3)∪(5,  ∞)\left( { - \infty ,\; - 5} \right) \cup \left( { - 3,\;3} \right) \cup \left( {5,\;\infty } \right)
Option C is the correct answer because the solution set is found by solving the inequality, considering the domain restrictions imposed by the absolute value and denominator.
9 Medium
9 of 914
Application +4 | -1
The least integral value of kk for which the quadratic expression (k−2)x2+8x+k+4(k-2) x^2+8 x+k+4 is positive for all real xx is
View answer and explanation
Correct option: C
5
Option C (5) is correct because it is the smallest integer value of k that satisfies both the discriminant condition and the leading coefficient condition for the quadratic to be positive.
10 Medium
10 of 914
Application +4 | -1
The value of xx for which x−34−x<x−12−x−23,2−x>2x−8\frac{x-3}{4}-x < \frac{x-1}{2}-\frac{x-2}{3}, 2-x > 2 x-8
View answer and explanation
Correct option: B
(−1,103)\left(-1, \frac{10}{3}\right)
Option B ((-1, 103\frac{{10}}{{3}})) is correct because solving the inequalities yields x>−1x > -1 and x<103x < \frac{{10}}{{3}}, thus the solution set is the interval between -1 and 103\frac{{10}}{{3}}.
11 Easy
11 of 914
Application +4 | -1
The number of real roots of the equation x2−3∣x∣+2=0x^2-3|x|+2=0
View answer and explanation
Correct option: A
4
Option A (4) is correct because substituting |x| = y results in a quadratic equation with two positive solutions for |x|, leading to four real solutions for x.
12 Medium
12 of 914
Application +4 | -1
Let pp and qq be two non-zero real numbers and α,β\alpha, \beta are two numbers such that α3+β3=−p,αβ=q\alpha^3+\beta^3=-p, \alpha \beta=q, then the quadratic equation whose roots are α2/β\alpha^2 / \beta and β2/α\beta^2 / \alpha is
View answer and explanation
Correct option: B
qx2+px+q2=0q x^2+p x+q^2=0
Option D (qx2+px+q2=0q x^2+p x+q^2=0) is correct because we can find the sum and product of the new roots in terms of p and q, and then construct the quadratic equation.
13 Medium
13 of 914
Analyzing +4 | -1
There is only one real value of ' aa ' for which the quadratic equation ax2+(a+3)x+a−3=0a x^2+(a+3) x+a-3=0 has two positive integral solutions. The product of these two solutions is
View answer and explanation
Correct option: B
8
The correct answer is B (8) because the product of the roots is found by solving for 'a' and substituting back into the product of roots formula.
14 Easy
14 of 914
Application +4 | -1
If z=x+iyz=x+i y, where x,y∈Rx, y \in R and i=−1i=\sqrt{-1} satisfies the relation ∣2z+5∣=∣6z−9∣|2 z+5|=|6 z-9|, then ∣z∣2=aRe⁡(z)+b|z|^2=a \operatorname{Re}(z)+b where a,b∈Ra, b \in R. The value of (a−4b)(a-4 b) is
View answer and explanation
Correct option: A
11
Option A (11) is correct because solving the given equation leads to ∣z∣2=4Re(z)−7/4|z|^2 = 4 Re(z) - 7/4, and then calculating a−4ba - 4b yields 11.
15 Easy
15 of 914
Analyzing +4 | -1
If α\alpha is a root of the equation 2x(2x+1)=1,2x\left( {2x + 1} \right) = 1,, then the other roots is
View answer and explanation
Correct option: C
4  α3−3  α4\;{\alpha ^3} - 3\;\alpha
Option C (4α3−3α4{\alpha ^3} - 3\alpha) is correct because by using the given equation and the relationship between the roots, the other root can be expressed in terms of α\alpha as 4α3−3α4{\alpha ^3} - 3\alpha.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
16 Easy
16 of 914
Analyzing +4 | -1
If x2+2x+n>10{x^2} + 2x + n > 10 for all real numbers xx, then which of the following conditions is true?
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Correct option: D
n>11n > 11
Option D (n>11n > 11) is correct because for the quadratic expression to be greater than 10 for all real x, the discriminant must be negative, leading to the condition n > 11.
17 Hard
17 of 914
Analyzing +4 | -1
If f(x)=∑k=2n(x−1k−1)(x−1k),f\left( x \right) = \sum \limits_{k = 2}^n \left( {x - \frac{1}{{k - 1}}} \right)\left( {x - \frac{1}{k}} \right), then the product of roots of f(x)=0f\left( x \right) = 0 as n→∞,n \to \infty , is
View answer and explanation
Correct option: B
0
Option B (0) is correct because as n approaches infinity, the product of the roots of the given function approaches 0.
18 Easy
18 of 914
Analyzing +4 | -1
If a,b,ca, b, c are real and a≠ba \neq b, the roots of the equation 2(a−b)x2−11(a+b+c)x−3(a−b)=02(a-b) x^2-11(a+b+c) x-3(a-b)=0 are
View answer and explanation
Correct option: B
real and unequal
Option B (real and unequal) is correct because the discriminant of the quadratic equation is positive, indicating two distinct real roots.
19 Medium
19 of 914
Analyzing +4 | -1
The equation 3x2+x+5=x−3\sqrt{3 x^{2}+x+5}=x-3, where xx is real, has;
View answer and explanation
Correct option: A
no solution
Option A (no solution) is correct because the solutions obtained after squaring do not satisfy the original equation.
20 Medium
20 of 914
Analyzing +4 | -1
Let α=−1+i32.\alpha = \frac{{ - 1 + i\sqrt 3 }}{2}. If a =(1+α)∑k=0100α2k  and  b = ∑k=0100α3k  ,a\, = (1 + \alpha )\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}\,\,\mathrm{and}\,\,b\, = \,} \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}\,\,} , then aa and bb the roots of the quadratic equation.
View answer and explanation
Correct option: C
x2−102x+101=0{x^2} - 102x + 101 = 0
Option C (x2−102x+101=0{x^2} - 102x + 101 = 0) is correct because it is the quadratic equation formed with roots a=1 and b=101, derived from the properties of the complex cube root of unity.
21 Medium
21 of 914
Application +4 | -1
Let α\alpha and β\beta be the roots of x2−6x−2=0x^2-6 x-2=0, with α>β\alpha > \beta. If an=αn−βna_n=\alpha^n-\beta^n for n≥1n \geq 1, then the value of a10−2a82a9\frac{a_{10}-2 a_8}{2 a_9} is
View answer and explanation
Correct option: A
3
Option A (3) is correct because using the recurrence relation derived from the quadratic equation, the expression simplifies to 3.
22 Medium
22 of 914
Analyzing +4 | -1
Let a=cos⁡2π7+isin⁡2  π7,α=a+a2+a4a = \cos \frac{{2\pi }}{7} + i\sin \frac{{2\;\pi }}{7},\alpha = a + {a^2} + {a^4} and β=a3+a5+a6.\beta = {a^3} + {a^5} + {a^6}. Then, the equation whose roots are α,β\alpha ,\beta is
View answer and explanation
Correct option: D
x2+x+2=0{x^2} + x + 2 = 0
Option D (x² + x + 2 = 0) is correct because the equation whose roots are α and β is found by calculating α + β and αβ and substituting them into the quadratic equation.
23 Easy
23 of 914
Application +4 | -1
The quadratic equation whose roots are sec⁡2  θ{\sec ^2}\,\,\theta and cosec2    θ\mathrm{cose}{\mathrm{c}^2}\,\,\;\theta can be
View answer and explanation
Correct option: C
x2−4x+4=0{x^2} - 4x + 4 = 0
Option C (x2−4x+4=0{x^2} - 4x + 4 = 0) is correct because it is a quadratic equation with roots sec⁡2 θ{\sec ^2}\,{\theta } and cosec2   θ\mathrm{cose}{\mathrm{c}^2}\,\;{\theta } and has a discriminant that is greater than or equal to 4.
24 Easy
24 of 914
Application +4 | -1
If roots of the equation x2+ax+b=0x^2+a x+b=0 are α,β\alpha, \beta, then the roots of x2+(2α+a)x+α2+x^2+(2 \alpha+a) x+\alpha^2+ aα+b=0a \alpha+b=0 are
View answer and explanation
Correct option: C
0,β−α0, \beta-\alpha
Option C (0, β−α\beta - \alpha) is correct. By substituting 0 into the second equation, we find it is a root, and the other root can be found using Vieta's formulas.
25 Easy
25 of 914
Analyzing +4 | -1
(a2−3a+2)x2+(a2−5a+6)x+a−2=r\left( {{a^2} - 3a + 2} \right){x^2} + \left( {{a^2} - 5a + 6} \right)x + a - 2 = r for three distinct values of xx for some r∈R,   ifr \in R,\;\,\mathrm{if} a+ra + r is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because when a = 2, the equation holds true for three distinct values of x, and r = 0, thus a + r = 2.
26 Easy
26 of 914
Understanding +4 | -1
If both the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0 are zero, then
View answer and explanation
Correct option: A
b=c=0b = c = 0
Option A (b=c=0b = c = 0) is correct because if both roots of the quadratic equation are zero, then the sum and product of the roots are zero, implying that b and c must be zero.
27 Easy
27 of 914
Application +4 | -1
If the roots of (a2+b2)x2−2(bc+ad)x+c2+d2=0\left( {{a^2} + {b^2}} \right){x^2} - 2\left( {bc + ad} \right)x + {c^2} + {d^2} = 0 are equal, then
View answer and explanation
Correct option: A
ab=cd\frac{a}{b} = \frac{c}{d}
Option A (ab=cd\frac{a}{b} = \frac{c}{d}) is correct because for the roots to be equal, the discriminant must be zero, which simplifies to ad = bc, hence a/b = c/d.
28 Easy
28 of 914
Application +4 | -1
If a≤0,a \le 0, then the real values of xx satisfying x2−2a∣x−a∣−3a2=0{x^2} - 2a\left| {x - a} \right| - 3{a^2} = 0 are
View answer and explanation
Correct option: A
a(1−2),a(−1+6)a\left( {1 - \sqrt 2 } \right),a\left( { - 1 + \sqrt 6 } \right)
Option A (a(1−2),a(−1+6)a\left( {{{{1 - \sqrt 2 }}}} \right),a\left( {{{{ - 1 + \sqrt 6 }}}} \right)) is correct because these are the real values of x that satisfy the given equation when considering both cases for the absolute value and the condition a <= 0.
29 Easy
29 of 914
Analyzing +4 | -1
If α,β\alpha, \beta are the roots of the equation
x2−(5+3log⁡35−5log⁡53)+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)+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}, is
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 after simplifying the given equation and finding the roots alpha and beta, the equation with roots alpha+1/beta and beta+1/alpha is derived correctly.
30 Easy
30 of 914
Analyzing +4 | -1
If α,β\alpha, \beta are the roots of (x−a)(x−b)=c,c≠0(x-a)(x-b)=c, c \neq 0, then the roots of (x−α)(x−β)+c=0(x-\alpha)(x-\beta)+c=0 shall be:
View answer and explanation
Correct option: C
a,ba, b
Option C (a,ba, b) is correct because the roots of the new equation are aa and bb, derived from manipulating the original equation and its roots.
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