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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 631-660 of 914 questions

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631 Easy
631 of 914
Application +4 | -1
The number of integral values of mm for which the equation (1+m2)x2−2(1+3m)x+(1+8m)=0,(1 + {m^2}){x^2} - 2(1 + 3m)x + (1 + 8m) = 0, has no real root is
View answer and explanation
Correct option: B
infinitely many
Option B (infinitely many) is correct because the discriminant of the quadratic equation must be negative, leading to the condition m > 0, which has infinitely many solutions.
632 Easy
632 of 914
Analyzing +4 | -1
The number of integral solutions of x+2x2+1>12\frac{{x + 2}}{{{x^2} + 1}} > \frac{1}{2} is
View answer and explanation
Correct option: C
3
Option C (3) is correct because there are three integers within the solution interval of the inequality.
633 Medium
633 of 914
Application +4 | -1
If the roots of the equation x2+px+q=0{x^2} + px + q = 0 are tan⁡300\tan {30^0} and tan⁡150\tan {15^0} respectively, then the value of 2+q−p2 + q - p is
View answer and explanation
Correct option: A
3
Option A (3) is correct because by using the sum and product of roots, and the tangent addition formula, we can find the value of the expression 2 + q - p.
634 Easy
634 of 914
Application +4 | -1
The roots of the equation (q−r)x2+(r−p)x+(p−q)=0\left( {q - r} \right){x^2} + \left( {r - p} \right)x + \left( {p - q} \right) = 0 are
View answer and explanation
Correct option: B
p−qq−r,  1\frac{{p - q}}{{q - r}},\;1
Option B p−qq−r,  1\frac{{p - q}}{{q - r}},\;1 is correct because these are the roots obtained by factoring the given quadratic equation.
635 Medium
635 of 914
Analyzing +4 | -1
If 0<c<a<b<d0< c < a < b < d, then roots of the equation bx2+{1−b(c+d)}x+bcd−a=0b x^2+\{1-b(c+d)\} x+b c d-a=0 are
View answer and explanation
Correct option: B
real and distinct in which one lie between cc and dd
Option B (real and distinct in which one lie between cc and dd) is correct because f(c)f(c) is negative and f(d)f(d) is positive, indicating a root between cc and dd.
636 Easy
636 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of the equation x2−ax+b=0{x^2} - ax + b = 0 and An=αn+βn,{A_n} = {\alpha ^n} + {\beta ^n}, then which one of the following is true?
View answer and explanation
Correct option: C
An+1=a  An−b  An−1{A_{n + 1}} = a\;{A_n} - b\;{A_{n - 1}}
Option C (An+1=a  An−b  An−1{A_{n + 1}} = a\;{A_n} - b\;{A_{n - 1}}) is correct because it represents the recurrence relation derived from the roots of the quadratic equation.
637 Easy
637 of 914
Analyzing +4 | -1
Let α\alpha and β\beta be the roots of x2−6x−2=0{x^2} - 6x - 2 = 0. If an=αn−βn{a_n} = {\alpha ^n} - {\beta ^n} for n≥1n \ge 1, then the value of a10−2a83a9\frac{{{a_{10}} - 2{a_8}}}{{3{a_9}}} is
View answer and explanation
Correct option: C
2
Option C (2) is correct because by substituting the roots into the given expression and using the quadratic equation, the expression simplifies to 2.
638 Easy
638 of 914
Application +4 | -1
Let (sin⁡a)x2+(sin⁡a)x+(1−cos⁡a=0)\left( {\sin a} \right){x^2} + \left( {\sin a} \right)x + \left( {1 - \cos a = 0} \right). The value of aa for which roots of this equation are real and distinct, is
View answer and explanation
Correct option: A
(0,2tan⁡−114)\left( {0,2{{\tan }^{ - 1}}\frac{1}{4}} \right)
Option A ((0,2tan⁡−114)(0,2{{\tan }^{ - 1}}\frac{1}{4})) is correct because the range of 'a' is determined by the discriminant of the quadratic equation being greater than zero.
639 Easy
639 of 914
Analyzing +4 | -1
sin⁡A,sin⁡B,cos⁡A\sin A,\sin B,\cos A are in GP. Roots of x2+2xcot⁡B+1=0  {x^2} + 2x\cot B + 1 = 0\; are always
View answer and explanation
Correct option: A
Real
Option A (Real) is correct because the discriminant of the quadratic equation is non-negative, implying that the roots are real.
640 Medium
640 of 914
Analyzing +4 | -1
If a(p+q)2+2b  pq+c=0a{\left( {p + q} \right)^2} + 2b\;pq + c = 0 and a(p+r)2+2b  pr+c=0,a{\left( {p + r} \right)^2} + 2b\;pr + c = 0, then qr=qr =
View answer and explanation
Correct option: A
p2+ca{p^2} + \frac{c}{a}
Option A (p2+ca{p^2} + \frac{c}{a}) is correct because it represents the product of the roots qq and rr of the quadratic equation derived from the given conditions.
641 Easy
641 of 914
Application +4 | -1
The value of λ\lambda for which the quadratic equation 3x2+2(λ2+1)x+(λ2−3λ+2)=03 x^2+2\left(\lambda^2+1\right) x+\left(\lambda^2-3 \lambda+2\right)=0 has roots of opposite signs, lies in the interval.
View answer and explanation
Correct option: A
(1,2)(1,2)
Option A ((1,2)(1,2)) is correct because the product of the roots of the quadratic equation must be negative, leading to the inequality (λ−1)(λ−2)<0(\lambda-1)(\lambda-2) < 0, which is satisfied when 1<λ<21 < \lambda < 2.
642 Medium
642 of 914
Application +4 | -1
The value of 'a' for which one root of quadratic equation (a2−5a+3)x2+(3a−1)x+2=0\left(a^2-5 a+3\right) x^2+(3 a-1) x+2=0 is twice as large as other is
View answer and explanation
Correct option: A
2/32 / 3
Option A (2/3) is correct because it satisfies the condition derived from the relationship between the roots of the quadratic equation.
643 Easy
643 of 914
Application +4 | -1
The solution set of inequation log⁡1/3(2x+2−4x)≥−2,{\log _{1/3}}\left( {{2^{x + 2}} - {4^x}} \right) \ge - 2, is
View answer and explanation
Correct option: C
(−∞,2)\left( { - \infty ,2} \right)
Option C ((−∞,2)\left( { - \infty ,2} \right)) is correct because solving the logarithmic inequality leads to the condition x<2x < 2.
644 Easy
644 of 914
Application +4 | -1
The roots of ax2+2  bx+c=0a{x^2} + 2\;bx + c = 0 and bx2−2ac  x+b=0b{x^2} - 2\sqrt {ac} \;x + b = 0 are simultaneously real, then
View answer and explanation
Correct option: B
ac=b2ac = {b^2}
Option B (ac=b2ac = {b^2}) is correct because for the roots to be simultaneously real, the discriminants of both quadratic equations must be non-negative, leading to this condition.
645 Medium
645 of 914
Understanding +4 | -1
If 0<a<b<c0 < a < b < c and the roots α\alpha and β\beta of the equation ax2+bx+c=0a x^2+b x+c=0 are imaginary, then
View answer and explanation
Correct option: A
∣α∣=∣β∣|\alpha|=|\beta|
Option A (∣α∣=∣β∣\left|\alpha\right|=\left|\beta\right|) is correct because when the roots of a quadratic equation are imaginary, they are complex conjugates, and complex conjugates have the same magnitude.
646 Easy
646 of 914
Application +4 | -1
The value of pp for which the quadratic equation x2−px+p+3=0x^2-p x+p+3=0 has reciprocal roots is
View answer and explanation
Correct option: D
-2
Option D (-2) is correct because for reciprocal roots, the product of the roots is 1, which allows us to solve for p.
647 Easy
647 of 914
Analyzing +4 | -1
In writing an equation of the form ax2+bx+c=0;a{x^2} + bx + c = 0; the coefficient of xx is written incorrectly and roots are found to be equal. Again in writing the same equation the constant term is written incorrectly and it is found that one root is equal to those of the previous wrong equation while the other is double of it. If α\alpha and β\beta be the roots of correct equation, then (α−β)2{\left( {\alpha - \beta } \right)^2} is equal to
View answer and explanation
Correct option: B
5 α  β5\,\alpha \;\beta
Option B (5,α  β5\\,\alpha \;\beta) is correct because by analyzing the incorrect equations and relating their roots to the correct equation's roots, we can derive the relationship (α−β)2=5αβ{\left( {\alpha - \beta } \right)^2} = 5\alpha \beta.
648 Medium
648 of 914
Application +4 | -1
If α,β\alpha, \beta are roots of x2−2x−1=0x^2-2 x-1=0, then value of 5α4+12β35 \alpha^4+12 \beta^3 is
View answer and explanation
Correct option: B
169
Option B (169) is correct because substituting the roots into the given equation and simplifying the expression 5α4+12β35 \alpha^4 + 12 \beta^3 yields 169.
649 Easy
649 of 914
Analyzing +4 | -1
If x1,x2,x3{x_1},{x_2},{x_3} are distinct roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0, then
View answer and explanation
Correct option: D
a=b=c=0a = b = c = 0
Option D (a=b=c=0a = b = c = 0) is correct because a quadratic equation cannot have three distinct roots unless it is an identity, which requires all coefficients to be zero.
650 Easy
650 of 914
Analyzing +4 | -1
The numbers of pairs (a,  b)(a,\,\,b) of real numbers, such that whenever α\alpha is a root of the equation x2+ax+b=0,   α2−2{x^2} + ax + b = 0,\,\,\,{\alpha ^2} - 2 is also a root of this equation, is
View answer and explanation
Correct option: A
6
The correct answer is A (6) because there are six possible pairs of (a, b) that satisfy the given condition regarding the roots of the quadratic equation.
651 Medium
651 of 914
Application +4 | -1
The real values of aa for which the quadratic equation 3x2+2(a2+1)x+(a2−3a+2)=03 x^2+2\left(a^2+1\right) x+\left(a^2-3 a+2\right)=0 possesses roots of opposite signs lie in
View answer and explanation
Correct option: C
(1,2)(1,2)
Option C ((1,2)) is correct because the product of roots must be negative, leading to the interval (1,2).
652 Medium
652 of 914
Application +4 | -1
If one root is square of the other root of the equation x2+px+q=0x^2+p x+q=0, then the relation between pp and qq is
View answer and explanation
Correct option: A
p3−(3p−1)q+q2=0p^3-(3 p-1) q+q^2=0
Option A (p3−(3p−1)q+q2=0p^3-(3 p-1) q+q^2=0) is correct because it represents the relationship between p and q when one root is the square of the other root of the given quadratic equation.
653 Medium
653 of 914
Application +4 | -1
If α,β\alpha ,\beta are the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0, then the equation whose roots are 2α+3β2\alpha + 3\beta and 3α+2β3\alpha + 2\beta, is
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct because the derived equation does not match any of the provided options.
654 Easy
654 of 914
Application +4 | -1
The coefficients of xx in the quadratic equation x2+bx+c=0{x^2} + bx + c = 0 was taken as 17 in place of 13, its roots were found to be −2- 2 and −15- 15. The correct root of the original equation are
View answer and explanation
Correct option: A
−10,−3- 10, - 3
Option A (−10,−3- 10, - 3) is correct because these are the correct roots of the original equation after finding the correct constant term.
655 Medium
655 of 914
Application +4 | -1
If both the roots of the equation x2+ax+b+1=0x^2+a x+b+1=0 are positive integers, then a possible value of a2+b2a^2+b^2 is
View answer and explanation
Correct option: A
50
Option A (50) is correct because it is a possible value of a2+b2a^2 + b^2 when both roots of the equation are positive integers.
656 Easy
656 of 914
Application +4 | -1
Let α\alpha and β\beta be the roots of equation px2+qx+r=0,   p≠0.p{x^2} + qx + r = 0,\,\,\,p \ne 0. If p,  qp,\,\,q and rr are in AP and 1α+1β=4,\frac{1}{\alpha } + \frac{1}{\beta } = 4, then the value of ∣α−β∣\left| {\alpha - \beta } \right| is
View answer and explanation
Correct option: D
2139\frac{{2\sqrt {13} }}{9}
Option D (2139\frac{{2\sqrt {13} }}{9}) is correct because by using the given conditions and properties of arithmetic progressions and quadratic equations, the value of ∣α−β∣\left| {\alpha - \beta } \right| is calculated to be 2139\frac{{2\sqrt {13} }}{9}.
657 Easy
657 of 914
Application +4 | -1
A solution of the set of equations x+y+z=xy+yz+zx=0x+y+z=x y+y z+z x=0, x3+y3+z3=−3x^3+y^3+z^3=-3 is-
View answer and explanation
Correct option: D
−1,−ω,−ω2-1,-\omega,-\omega^2
Option D, −1,−ω,−ω2-1,-\omega,-\omega^2, is the correct solution because it satisfies all three given equations.
658 Easy
658 of 914
Application +4 | -1
Let f(x)f\left( x \right) be a quadratic polynomial such that f(−1)+f(2)=0.f\left( { - 1} \right) + f\left( 2 \right) = 0. If one of the roots of f(x)=0f\left( x \right) = 0 is 3, then its other root lies in
View answer and explanation
Correct option: A
(−1,0)\left( { - 1,0} \right)
The correct answer is A ((-1,0)) because it is the interval in which the other root of the quadratic polynomial lies, determined by using the given condition f(-1) + f(2) = 0.
659 Medium
659 of 914
Application +4 | -1
If the sum of the greatest integer less than or equal to xx and the least integer greater than or equal to xx is 5, then the solution set for xx is
View answer and explanation
Correct option: A
(2,3)\left( {2,3} \right)
Option A ((2,3)\left( {2,3} \right)) is correct because the sum of the greatest integer less than or equal to xx and the least integer greater than or equal to xx is 5 only when 2<x<32 < x < 3.
660 Easy
660 of 914
Application +4 | -1
If the roots of the equation ax2+bx+c=0   be   α  and  β  ,a{x^2} + bx + c = 0\;\,\mathrm{be}\,\;\alpha \;\mathrm{and}\;\beta \;, then the roots of the equation cx2+bx+a=0c{x^2} + bx + a = 0 are
View answer and explanation
Correct option: C
1α,1β\frac{1}{\alpha },\frac{1}{\beta }
Option C (1α,1β\frac{1}{\alpha },\frac{1}{\beta }) is correct because the roots of the new equation are the reciprocals of the roots of the original equation.
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