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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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Showing 601-630 of 914 questions

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601 Medium
601 of 914
Application +4 | -1
If   sec⁡  α    and    cosec  α  \mathrm{\;}\sec \,\,\mathrm{\alpha \;}\,\,\mathrm{and\;}\,\,\mathrm{cosec}\,\,\mathrm{\alpha \;} are the root of the equation x2−px+q=0{x^2} - px + q = 0,then
View answer and explanation
Correct option: B
p2=q(q+2){p^2} = q\left( {q + 2} \right)
Option A (1\mathrm{1}) is correct because by substituting sin⁡x=cos⁡2x\sin x = {\cos ^2}x into the expression and simplifying, we obtain 1.
602 Easy
602 of 914
Application +4 | -1
Let α,β\alpha ,\beta be the roots of x2+bx+1=0.{x^2} + bx + 1 = 0. Then the equation whose roots are −(α+1β)- \left( {\alpha + \frac{1}{\beta }} \right) and −(β+1α),- \left( {\beta + \frac{1}{\alpha }} \right), is
View answer and explanation
Correct option: C
x2−2    b  x+4=0{x^2} - 2\;\;b\;x + 4 = 0
Option C (x2−2    b  x+4=0{x^2} - 2\;\;b\;x + 4 = 0) is correct because it is formed by finding the sum and product of the new roots and using them to construct the quadratic equation.
603 Easy
603 of 914
Application +4 | -1
Suppose that two persons AA and BB solve the equation x2+ax+b=0{x^2} + ax + b = 0. While solving AA commits a mistake in the coefficient of xx was taken as 15 in place of −9- 9 and finds the roots as −7- 7 and −2- 2. Then the equation is
View answer and explanation
Correct option: B
x2−9x+14=0{x^2} - 9x + 14 = 0
Option B (x2−9x+14=0{x^2} - 9x + 14 = 0) is correct because it uses the correct coefficient of the constant term (product of roots) obtained from the incorrect equation solved by person A.
604 Easy
604 of 914
Application +4 | -1
If the equations x2+ax+b=0x^{2}+a x+b=0 and x2+a′x+b′=0x^{2}+a^{\prime} x+b^{\prime}=0 have a common root, then this common root is equal to
View answer and explanation
Correct option: D
(b−b′)(a′−a)\frac{\left(b-b^{\prime}\right)}{\left(a^{\prime}-a\right)}
Option D ((b−b′)(a′−a)\frac{{\left(b-b^{\prime}\right)}}{{\left(a^{\prime}-a\right)}}) is correct because by subtracting the two equations, the common root can be isolated and solved for.
605 Medium
605 of 914
Analyzing +4 | -1
If x1,x2,x3,…,xnx_1, x_2, x_3, \ldots, x_n are the roots of xn+ax+b=0x^n+a x+b=0, then the value of (x1−x2)(x1−x3)(x1−x4)…(x1−xn)\left(x_1-x_2\right)\left(x_1-x_3\right)\left(x_1-x_4\right) \ldots\left(x_1-x_n\right) is
View answer and explanation
Correct option: C
n(x1)n−1+an\left(x_1\right)^{n-1}+a
Option C (n(x1)n−1+an\left(x_1\right)^{n-1}+a) is correct because it is the result of differentiating the polynomial and evaluating it at x1x_1.
606 Easy
606 of 914
Analyzing +4 | -1
The value of 'kk' for which one of the roots of x2−x+3  k=0,{x^2} - x + 3\;k = 0, is double of one of the roots of x2−x+k=0{x^2} - x + k = 0 is
View answer and explanation
Correct option: B
−2- 2
Option B (−2- 2) is correct because by setting up the relationship between the roots of the two quadratic equations and solving the resulting equations, we find that k = -2.
607 Easy
607 of 914
Application +4 | -1
If α,β\alpha, \beta are roots of x2−3x+1=0x^2-3 x+1=0, then the equation whose roots are 1α−2,1β−2\frac{1}{\alpha-2}, \frac{1}{\beta-2} is:
View answer and explanation
Correct option: C
x2−x−1=0 dx^2-x-1=0 \mathrm{~d}
Option C (x2−x−1=0 dx^2-x-1=0 \mathrm{~d}) is correct because it is obtained by finding the sum and product of the new roots (1α−2\frac{1}{\alpha-2} and 1β−2\frac{1}{\beta-2}) and constructing the quadratic equation.
608 Medium
608 of 914
Analyzing +4 | -1
If cos⁡4θ+α,sin⁡4θ+α\cos ^4 \theta+\alpha, \sin ^4 \theta+\alpha are the roots of the equation x2+2bx+b=0x^2+2 b x+b=0 and cos⁡2θ+β,sin⁡2θ+β\cos ^2 \theta+\beta, \sin ^2 \theta+\beta are the roots of x2+4x+2=0x^2+4 x+2=0 the value of b{b} is
View answer and explanation
Correct option: C
−1-1
Option C (-1) is correct because the difference of roots of the given equations is the same, leading to a quadratic equation in 'b' which gives b = -1 as a valid solution.
609 Easy
609 of 914
Application +4 | -1
The value of
x=4+15+14+15+14+...∞x = 4 + \frac{1}{{5 + \frac{1}{{4 + \frac{1}{{5 + \frac{1}{{4 + ...\infty }}}}}}}}
View answer and explanation
Correct option: A
2+25302 + \frac{2}{5}\sqrt {30}
Option A (2+25302 + \frac{2}{5}\sqrt {30}) is correct because it is the positive solution to the quadratic equation derived from the continued fraction.
610 Easy
610 of 914
Analyzing +4 | -1
If a<b<c<da < b < c < d, then the roots of the equation (x−a)(x−c)+2(x−b)(x−d)=0\left( {x - a} \right)\left( {x - c} \right) + 2\left( {x - b} \right)\left( {x - d} \right) = 0 are
View answer and explanation
Correct option: A
Real and distinct
Option A (Real and distinct) is correct because the discriminant of the given equation is positive, indicating that the roots are real and distinct.
611 Medium
611 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of the equation (5+2)x2−(5+\sqrt{2}) x^2- (4+5)x+8+25=0(4+\sqrt{5}) x+8+2 \sqrt{5}=0, then 2(1/α)+(1/β)\frac{2}{(1 / \alpha)+(1 / \beta)} is equal to
View answer and explanation
Correct option: B
4
Option B (4) is correct because it is calculated by using Vieta's formulas to find the sum and product of the roots and then simplifying the given expression.
612 Easy
612 of 914
Application +4 | -1
Two students while solving a quadratic equation in xx, one copied the constant term incorrectly and got the roots 3 and 2. The other copied the constant term coefficient of x2{x^2} correctly as −6- 6 and 1 respectively the correct roots are
View answer and explanation
Correct option: D
6,  −16,\; - 1
Option D (6,  −16,\; - 1) is correct because by using the information about the incorrect roots, we can determine the correct sum and product of the roots, leading to the correct roots of 6 and -1.
613 Medium
613 of 914
Analyzing +4 | -1
If the roots of a1x2+b1x+c1=0a_1 x^2+b_1 x+c_1=0 are α1,β1\alpha_1, \beta_1 and that of a2x2+b2x+c2=0a_2 x^2+b_2 x+c_2=0 are α2\alpha_2, β2\beta_2 such that α1α2=β1β2=1\alpha_1 \alpha_2=\beta_1 \beta_2=1, then
View answer and explanation
Correct option: B
a1c2=b1b2=c1a2\frac{a_1}{c_2}=\frac{b_1}{b_2}=\frac{c_1}{a_2}
Option B is correct because the roots of the second equation are reciprocals of the roots of the first equation.
614 Easy
614 of 914
Application +4 | -1
The sum of all values of xx, so that 16(x2+3x−1)=8(x2+3x+2)16^{\left(x^2+3 x-1\right)}=8^{\left(x^2+3 x+2\right)}, is
View answer and explanation
Correct option: C
-3
Option C, -3, is correct because solving the exponential equation 16(x2+3x−1)=8(x2+3x+2)16^{(x^2+3x-1)} = 8^{(x^2+3x+2)} leads to a quadratic equation whose roots sum to -3.
615 Medium
615 of 914
Analyzing +4 | -1
If x2+6x−27>0x^2+6 x-27 > 0 and x2−3x−4<0x^2-3 x-4 < 0, then
View answer and explanation
Correct option: C
3<x<43 < x < 4
Option C (3<x<43 < x < 4) is correct because it represents the intersection of the solution sets for the two given inequalities.
616 Easy
616 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of the equation x2+px+q=0x^2+p x+q=0, then the value of (ωα+ω2β)(ω2α+ωβ)\left(\omega \alpha+\omega^2 \beta\right)\left(\omega^2 \alpha+\omega \beta\right), where ω\omega is an imaginary cube root of unity, is
View answer and explanation
Correct option: B
p2−3qp^2-3 q
Option B (p2−3qp^2-3 q) is correct because it is the simplified expression of (ωα+ω2β)(ω2α+ωβ)(\omega \alpha+\omega^2 \beta)(\omega^2 \alpha+\omega \beta) using the properties of cube roots of unity and Vieta's formulas.
617 Easy
617 of 914
Application +4 | -1
The value of x=2+2+2+……x=\sqrt{2+\sqrt{2+\sqrt{2+\ldots \ldots}}} is;
View answer and explanation
Correct option: B
2
Option B (2) is correct because the value is found by setting up an equation x = sqrt(2+x) and solving the quadratic equation.
618 Medium
618 of 914
Analyzing +4 | -1
The number of solutions of the equation (1+ex2)1+x21+x4−x2=1+cos⁡x,\frac{{\left( {1 + {e^{{x^2}}}} \right)\sqrt {1 + {x^2}} }}{{\sqrt {1 + {x^4}} - {x^2}}} = 1 + \cos x, is
View answer and explanation
Correct option: A
1
The correct answer is A. By analyzing the LHS and RHS of the equation, we find that the only possible solution is x=0.
619 Medium
619 of 914
Application +4 | -1
If the least value of the expression y=x2−ax+ay=x^2-a x+a is 1; then the value of aa is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because the least value of the quadratic expression is found by completing the square or using the vertex formula, leading to a = 2.
620 Easy
620 of 914
Analyzing +4 | -1
If α\alpha and β\beta are the distinct roots of the equation x2+(3)1/4x+31/2=0{x^2} + {(3)^{1/4}}x + {3^{1/2}} = 0 then the value of α96(α12−1)+β96(β12−1){\alpha ^{96}}\left( {{\alpha ^{12}} - 1} \right) + {\beta ^{96}}\left( {{\beta ^{12}} - 1} \right) is equal to
View answer and explanation
Correct option: C
52×32452 \times {3^{24}}
The correct answer is C (52×32452 \times {3^{24}}) because it results from simplifying the expression after substituting the roots of the given quadratic equation.
621 Hard
621 of 914
Analyzing +4 | -1
Let [x]\left[ x \right] denote the greatest integer less than or equal to xx. Then, in [0,3]\left[ {0,3} \right] the number of solutions of the equation x2−3x+[x]=0,{x^2} - 3x + \left[ x \right] = 0, is
View answer and explanation
Correct option: C
2
Option C (2) is correct because by analyzing the equation in different intervals of [x], we find two solutions.
622 Easy
622 of 914
Application +4 | -1
If the equation 5x2−10x+log⁡1/5a=05 x^{2}-10 x+\log _{1 / 5} a=0 has real roots, then the minimum value of aa is
View answer and explanation
Correct option: A
155\frac{1}{5^{5}}
Option A (155\frac{{1}}{{5^{5}}}) is correct because the equation having real roots implies that the discriminant is non-negative, leading to the minimum value of aa.
623 Easy
623 of 914
Application +4 | -1
Let f(x)=∫1x(2−t2)dtf(x)=\int_1^x \sqrt{\left(2-t^2\right)} d t, the real roots of the equation x2−f′(x)=0x^2-f^{\prime}(x)=0 are
View answer and explanation
Correct option: A
±1\pm 1
Option A ±1\pm 1 is correct because these are the real roots obtained by solving the equation x2−f′(x)=0x^2 - f'(x) = 0 after finding f′(x)f'(x).
624 Medium
624 of 914
Application +4 | -1
If a,ba, b and cc are the sides of a triangle, then ab+c−a+bc+a−b+ca+b−c\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}
View answer and explanation
Correct option: B
≥3\geq 3
The correct answer is B (≥3\geq 3) because it is proven by substituting x=b+c−a,y=c+a−b,z=a+b−cx=b+c-a, y=c+a-b, z=a+b-c and applying the AM-GM inequality.
625 Easy
625 of 914
Application +4 | -1
If a,b∈R,a≠0a, b \in R, a \neq 0 and the quadratic equation ax2−bx+1=0a x^{2}-b x+1=0 has imaginary roots, then a+b+1a+b+1 is
View answer and explanation
Correct option: A
positive
Option A (positive) is correct because the quadratic equation having imaginary roots implies the function is always positive, and therefore f(-1) is positive.
626 Easy
626 of 914
Application +4 | -1
The product of the roots of the equation xlog⁡10x=104x3x^{\log _{10} x}=\frac{10^4}{x^3}, is
View answer and explanation
Correct option: C
11000\frac{1}{1000}
Option C (11000\frac{{1}}{{1000}}) is correct because it represents the product of the roots of the given logarithmic equation.
627 Easy
627 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 roots of

ax2−bx(x−1)+c(x−1)2=0a x^2-b x(x-1)+c(x-1)^2=0 (1)

are
View answer and explanation
Correct option: C
α1+α,β1+β\frac{\alpha}{1+\alpha}, \frac{\beta}{1+\beta}
Option C (α1+α,β1+β\frac{\alpha}{1+\alpha}, \frac{\beta}{1+\beta}) is correct because it is derived by substituting −xx−1-\frac{x}{x-1} into the original equation and solving for x.
628 Medium
628 of 914
Application +4 | -1
The number of solutions of the equation ∣x∣2−2∣x∣−8=0|x|^2-2|x|-8=0 which belong to the domain of the function f(x)=5−2xf(x)=\sqrt{5-2 x} is
View answer and explanation
Correct option: B
1
Option B (1) is correct because only one solution of the quadratic equation satisfies the domain restriction of the square root function.
629 Easy
629 of 914
Application +4 | -1
A real root of the equation log⁡4{log⁡2(x+8−x)}=0\log _4\left\{\log _2(\sqrt{x+8}-\sqrt{x})\right\}=0 is:
View answer and explanation
Correct option: A
1
Option A (1) is correct because by solving the logarithmic equation step-by-step, we find that the real root is x=1x=1.
630 Medium
630 of 914
Application +4 | -1
If one root of the equation x2+px+q=0x^2+p x+q=0 is the square of the other, then:
View answer and explanation
Correct option: D
p3+q2+q(1−3p)=0p^3+q^2+q(1-3 p)=0
Option D (p3+q2+q(1−3p)=0p^3+q^2+q(1-3 p)=0) is correct because by expressing the relationship between the roots and coefficients and manipulating the equations, we can derive the given relation.
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