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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Medium

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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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Showing 1-30 of 362 questions
1 Medium
1 of 362
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.
2 Medium
2 of 362
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.
3 Medium
3 of 362
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.
4 Medium
4 of 362
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.
5 Medium
5 of 362
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}}.
6 Medium
6 of 362
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.
7 Medium
7 of 362
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.
8 Medium
8 of 362
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.
9 Medium
9 of 362
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.
10 Medium
10 of 362
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.
11 Medium
11 of 362
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.
12 Medium
12 of 362
Application +4 | -1
The number of real values of xx satisfying the equation is

2(x2+1x2)−9(x+1x)+14=0   …(1)2\left(x^2+\frac{1}{x^2}\right)-9\left(x+\frac{1}{x}\right)+14=0\,\,\,…(1)
is
View answer and explanation
Correct option: C
3
Option C (3) is correct because there are three real values of x that satisfy the given equation after substitution and simplification.
13 Medium
13 of 362
Application +4 | -1
If α\alpha and β\beta are the roots of the equation 2x2−3x−6=02 x^2-3 x-6=0, then the equation whose roots are α2+2\alpha^2+2 and β2+2\beta^2+2 is
View answer and explanation
Correct option: C
4x2−49x+118=04 x^2-49 x+118=0
Option C (4x2−49x+118=04 x^2-49 x+118=0) is correct because it represents the quadratic equation with roots α2+2\alpha^2+2 and β2+2\beta^2+2, derived from the original equation's roots α\alpha and β\beta.
14 Medium
14 of 362
Application +4 | -1
If one real root of the quadratic equation 81x2+kx+256=081{x^2} + kx + 256 = 0 is cube of the other root, then a value of k is
View answer and explanation
Correct option: D
-300
Option D (-300) is correct because it's the value of k derived from the relationship between the roots of the quadratic equation.
15 Medium
15 of 362
Analyzing +4 | -1
The solution set for (2+2)x+(2−2)x=2⋅2x/4(\sqrt{2+\sqrt{2}})^x+(\sqrt{2-\sqrt{2}})^x=2 \cdot 2^{x / 4} is
View answer and explanation
Correct option: C
{0}\{0\}
Option C ({0}) is correct because x=0 is the only solution that satisfies the given equation, derived from the AM-GM inequality.

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16 Medium
16 of 362
Analyzing +4 | -1
It is given that P(x)=x3+ax2+bx+cP(x)=x^3+a x^2+b x+c has three distinct positive integral roots and P(34)=33P(34)=33. Let Q(x)=x2−2x+34Q(x)=x^2-2 x+34. If P(Q(x))P(Q(x)) has no real roots then cc is equal to
View answer and explanation
Correct option: D
-23529
Option D (-23529) is correct because by factoring P(x) and using P(34) = 33, we find the roots and then calculate c as the negative product of the roots.
17 Medium
17 of 362
Application +4 | -1
If 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 - \left( {a + 1} \right) = 0 then the value of aa is
View answer and explanation
Correct option: D
1
Option D (1) is correct because the sum of the squares of the roots is minimized when a=1, as derived from completing the square.
18 Medium
18 of 362
Analyzing +4 | -1
If the equation ax2+2  bx−3  c=0a{x^2} + 2\;bx - 3\;c = 0 has no real roots and 3  c4<a+b,\frac{{3\;c}}{4} < a + b, then
View answer and explanation
Correct option: A
c<0c < 0
Option A (c<0c < 0) is correct because the equation has no real roots, implying the quadratic is always positive or always negative, and given the condition, it must be always positive, leading to c<0c < 0.
19 Medium
19 of 362
Analyzing +4 | -1
If α\alpha and β\beta are the roots of 6x2−6x+1=06 x^2-6 x+1=0, then the value of 12[a+bα+cα2+dα3]+12[a+bβ+cβ2+dβ3]\frac{1}{2}\left[a+b \alpha+c \alpha^2+d \alpha^3\right]+\frac{1}{2}\left[a+b \beta+c \beta^2+d \beta^3\right] is:
View answer and explanation
Correct option: B
a1+b2+c3+d4\frac{a}{1}+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}
Option B (a1+b2+c3+d4\frac{a}{1}+\frac{b}{2}+\frac{c}{3}+\frac{d}{4}) is correct because it is the simplified form of the given expression after substituting the values of the sum and product of the roots.
20 Medium
20 of 362
Analyzing +4 | -1
If x2−(a−3)x+a=0x^2-(a-3) x+a=0 has atleast one positive root, then
View answer and explanation
Correct option: C
a∈(−∞,0]∪[9,∞)a \in(-\infty, 0] \cup[9, \infty)
Option C (a∈(−∞,0]∪[9,∞)a \in(-\infty, 0] \cup[9, \infty)) is correct because it includes all values of 'a' for which the quadratic equation has at least one positive root, considering both cases where one or both roots are positive.
21 Medium
21 of 362
Analyzing +4 | -1
The solution set of the inequation 5(1/4)(log⁡5x)2  ≥5  x(1/5)(log⁡5x){5^{\left( {1/4} \right){{\left( {{{\log }_5}x} \right)}^2}\;}} \ge 5\;{x^{\left( {1/5} \right)\left( {{{\log }_5}x} \right)}}, is
View answer and explanation
Correct option: C
(0,5−25]∪[525,∞)\left( {0,{5^{ - 2\sqrt 5 }}\left] \cup \right[{5^{2\sqrt 5 }},\infty } \right)
Option A ((2−a)(2−b)(2−c)≥8abc\left( {2 - a} \right)\left( {2 - b} \right)\left( {2 - c} \right) \ge 8abc) is correct because it is derived from applying the AM-GM inequality to the given conditions.
22 Medium
22 of 362
Analyzing +4 | -1
Number of solutions for x2−2−2[x]=0x^2-2-2[x]=0, (where, [] denotes the greatest integer function) is
View answer and explanation
Correct option: B
1
Option B (1) is correct because only one solution, x=6x = \sqrt{6}, satisfies the given equation and the greatest integer function condition.
23 Medium
23 of 362
Analyzing +4 | -1
If α\alpha and β\beta are the roots of the equation x2+px+q=0{x^2} + px + q = 0 and if the sum (α+β)x−α2+β22.x2+α3+β33.x3−...\left( {\alpha + \beta } \right)x - \frac{{{\alpha ^2} + {\beta ^2}}}{2}.{x^2} + \frac{{{\alpha ^3} + {\beta ^3}}}{3}.{x^3} - ... exists then it is equal to
View answer and explanation
Correct option: D
log⁡(1−px+qx2)\log \left( {1 - px + q{x^2}} \right)
Option D (log⁡(1−px+qx2)\log \left( {1 - px + q{x^2}} \right)) is correct because the given series can be expressed as the sum of two logarithmic series, which simplifies to the logarithm of a quadratic expression involving p, q, and x.
24 Medium
24 of 362
Analyzing +4 | -1
The set of all a∈IRa \in I R for which the equation x∣x−1∣+∣x+2∣+a=0x|x-1|+|x+2|+a=0 has exactly one real root, is
View answer and explanation
Correct option: B
(−∞,∞)(-\infty, \infty)
The correct answer is B ((−∞,∞)(-\infty, \infty)) because for any value of 'a', the equation will always have exactly one real root due to the nature of the piecewise function.
25 Medium
25 of 362
Analyzing +4 | -1
Let α,β\alpha ,\beta be the roots of the equation ax2+2bx+c=0a{x^2} + 2bx + c = 0 and γ,δ\gamma ,\delta be the roots of the equation px2+2qx+r=0p{x^2} + 2qx + r = 0. If α,β,γ,δ\alpha ,\beta ,\gamma ,\delta are in GP, then
View answer and explanation
Correct option: A
q2ac=b2pr{q^2}ac = {b^2}pr
Option A (q2ac=b2pr{q^2}ac = {b^2}pr) is correct because the relationship between the roots of the two quadratic equations, when they are in GP, leads to this condition.
26 Medium
26 of 362
Application +4 | -1
If roots of the equation (a2+b2)x2−2(ac+bd)x+c2+d2=0\left(a^2+b^2\right) x^2-2(a c+b d) x+c^2+d^2=0 are equal then :
View answer and explanation
Correct option: D
ad=bca d=b c
Option D (ad=bca d=b c) is correct because it is derived from the condition that the roots of the quadratic equation are equal, which implies the discriminant is zero.
27 Medium
27 of 362
Analyzing +4 | -1
The sum of the roots of the equation, x2+∣2x−3∣−4=0\mathrm{x}^{2}+|2 \mathrm{x}-3|-4=0, is:
View answer and explanation
Correct option: C
2\sqrt{2}
The sum of the roots of the equation is 2\sqrt{2} because we need to consider two cases for the absolute value and then check which roots satisfy the initial conditions.
28 Medium
28 of 362
Application +4 | -1
If α\alpha and β\beta are the roots of the equation x2−6x+a=0{x^2} - 6x + a = 0 and satisfy the relation 3α+2β=163\alpha + 2\beta = 16, then the value of aa is
View answer and explanation
Correct option: B
8
The correct answer is B (8) because by using the given relation 3α+2β=163\alpha + 2\beta = 16 and the properties of roots of a quadratic equation, we can solve for the value of 'a'.
29 Medium
29 of 362
Application +4 | -1
If α\alpha and β\beta are the roots of x2−2x+4=0{x^2} - 2x + 4 = 0, then the value of α6+β6{\alpha ^6} + {\beta ^6} is
View answer and explanation
Correct option: C
128
Option C (128) is correct because by finding the roots α\alpha and β\beta of the quadratic equation and substituting them into the expression α6+β6\alpha^6 + \beta^6, we get 128.
30 Medium
30 of 362
Application +4 | -1
If an angle AA of a △ABC\triangle A B C satisfies 5cos⁡A+3=05 \cos A+3=0, then the roots of the quadratic equation, 9x2+27x+20=09 x^{2}+27 x+20=0 are.
View answer and explanation
Correct option: B
sec⁡A,tan⁡A\sec A, \tan A
Option B (sec⁡A,tan⁡A\sec A, \tan A) is correct because the roots of the quadratic equation are equal to the secant and tangent of angle A, derived from the given cosine value.
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  • ✓ This set contains 362 single-correct-answer MCQs on QUADRATIC EQUATIONS.
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  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
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