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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Easy

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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 528 questions
1 Easy
1 of 528
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 528
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 528
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 Easy
4 of 528
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.
5 Easy
5 of 528
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.
6 Easy
6 of 528
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.
7 Easy
7 of 528
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.
8 Easy
8 of 528
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?
View answer and explanation
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.
9 Easy
9 of 528
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.
10 Easy
10 of 528
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.
11 Easy
11 of 528
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.
12 Easy
12 of 528
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.
13 Easy
13 of 528
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.
14 Easy
14 of 528
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.
15 Easy
15 of 528
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.

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16 Easy
16 of 528
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.
17 Easy
17 of 528
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.
18 Easy
18 of 528
Application +4 | -1
If the sum of the roots of the equation (a+1)x2+(2a+3)x+(3a+4)=0\left( {a + 1} \right){x^2} + \left( {2a + 3} \right)x + \left( {3a + 4} \right) = 0 is −1- 1, then the product of the roots is
View answer and explanation
Correct option: C
2
Option C (2) is correct because by using the given information about the sum of the roots and Vieta's formulas, we can find the value of 'a' and then calculate the product of the roots.
19 Easy
19 of 528
Application +4 | -1
The number real of roots of x−3(x2+7x+10)=0\sqrt{x-3}\left(x^2+7 x+10\right)=0, where x∈Rx \in \mathbf{R} is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the square root term requires x to be greater than or equal to 3, and only x=3 satisfies the equation.
20 Easy
20 of 528
Application +4 | -1
If one of the roots of the equation x2+bx+3=0{x^2} + bx + 3 = 0 is thrice the other, then bb is equal to
View answer and explanation
Correct option: D
±4\pm 4
Option D (±4\pm 4) is correct because by setting up the relationship between the roots and using Vieta's formulas, we can solve for the value of b.
21 Easy
21 of 528
Application +4 | -1
If the line x−23=y+15=z−2+2\frac{x-2}{3}=\frac{y+1}{5}=\frac{z-2}{+2} lies in the plane αx+2y+z−β\alpha x+2 y+z-\beta =0=0, then (α,β)(\alpha, \beta) equals
View answer and explanation
Correct option: C
(−4,−8)(-4,-8)
Option C ((-4,-8)) is correct because substituting the point (2,-1,2) into the plane equation and using the direction ratios of the line to satisfy the plane condition yields α = -4 and β = -8.
22 Easy
22 of 528
Application +4 | -1
if one of the roots of the equation x2+(1−3i)x−2(1+i)=0{x^2} + \left( {1 - 3i} \right)x - 2\left( {1 + i} \right) = 0 is −1+i,- 1 + i, then the other root is
View answer and explanation
Correct option: D
2i2i
Option D (2i2i) is correct because it is the other root of the quadratic equation, found using the sum of roots property.
23 Easy
23 of 528
Application +4 | -1
If xx is real, then the least value of the expression (ax2+bx+c),a>0\left(a x^2+b x+c\right), a>0 is
View answer and explanation
Correct option: C
4ac−b24a\frac{4 a c-b^2}{4 a}
Option C (4ac−b24a\frac{{4 a c-b^2}}{{4 a}}) is correct because it represents the minimum value of the quadratic expression, derived by completing the square.
24 Easy
24 of 528
Analyzing +4 | -1
Let SS denote the set of all values of SS for which the equation 2x2−2(2a+1)x+a(a+1)=02{x^2} - 2\left( {2a + 1} \right)x + a\left( {a + 1} \right) = 0 has one root less than aa and other root greater than aa, then SS equals
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the correct range for 'a' is a < -1 or a > 0, which is not represented by any of the given options.
25 Easy
25 of 528
Application +4 | -1
If α,β\alpha ,\beta are roots of x2+px−q=0{x^2} + px - q = 0 and γ,δ\gamma ,\delta are root of x2+px+r=0,{x^2} + px + r = 0, then the value of (α−γ)(α−δ)\left( {\alpha - \gamma } \right)\left( {\alpha - \delta } \right) is
View answer and explanation
Correct option: D
q+rq + r
Option D (q+rq + r) is correct because substituting the roots into the given equations and simplifying leads to this result.
26 Easy
26 of 528
Analyzing +4 | -1
If the roots of the equation bx2+cx+a=0b x^2+c x+a=0 be imaginary, then for all real values of xx, the expression 3b2x2+6bcx+2c23 b^2 x^2+6 b c x+2 c^2, is
View answer and explanation
Correct option: D
greater than (−4ab)(-4 a b)
Option D (greater than (−4ab)(-4 a b)) is correct because the minimum value of the expression is greater than −c2-c^2, and since the roots of the quadratic are imaginary, −c2>−4ab-c^2 > -4ab.
27 Easy
27 of 528
Application +4 | -1
The number of real solutions of the equation x2−∣x∣−12=0{x^2} - |x| - 12 = 0 is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the equation simplifies to an absolute value equation, yielding two real solutions.
28 Easy
28 of 528
Application +4 | -1
Given that, for all x∈Rx \in R, the expression x2−2x+4x2+2x+4\frac{x^2-2 x+4}{x^2+2 x+4} lies between 13\frac{1}{3} and 3, the values between which the expression 9⋅32x+6⋅3x+49⋅32x−6⋅3x+4\frac{9 \cdot 3^{2 x}+6 \cdot 3^x+4}{9 \cdot 3^{2 x}-6 \cdot 3^x+4} lies, are
View answer and explanation
Correct option: B
32\frac{3}{2} and 22
The correct answer is B (3/2 and 2) because by substituting t=3x+1t = 3^{x+1} and using the given range for the expression, we can find the range for the new expression.
29 Easy
29 of 528
Application +4 | -1
If a,b,ca,b,c are all positive and in H.P., then the roots of ax2+2  bx+c=0a{x^2} + 2\;bx + c = 0 are
View answer and explanation
Correct option: B
Imaginary
Option B (Imaginary) is correct because the discriminant of the quadratic equation is negative, indicating imaginary roots.
30 Easy
30 of 528
Analyzing +4 | -1
If the roots of the quadratic equation x−mmx+1=x+nnx+1\frac{x-m}{m x+1}=\frac{x+n}{n x+1} are reciprocal to each other, then:
View answer and explanation
Correct option: A
n=0n=0
Option A (n=0n=0) is correct because when the roots are reciprocal, their product is 1, leading to the equation m−n=m+nm-n = m+n, which simplifies to n=0n=0.
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