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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 361-390 of 528 questions

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361 Easy
361 of 528
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.
362 Easy
362 of 528
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.
363 Easy
363 of 528
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.
364 Easy
364 of 528
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.
365 Easy
365 of 528
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.
366 Easy
366 of 528
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.
367 Easy
367 of 528
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.
368 Easy
368 of 528
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.
369 Easy
369 of 528
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.
370 Easy
370 of 528
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.
371 Easy
371 of 528
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.
372 Easy
372 of 528
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.
373 Easy
373 of 528
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.
374 Easy
374 of 528
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}.
375 Easy
375 of 528
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.
376 Easy
376 of 528
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.
377 Easy
377 of 528
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.
378 Easy
378 of 528
Application +4 | -1
The quadratic equation whose roots are cubes of the roots of x2+bx+c=0x^2+b x+c=0 is
View answer and explanation
Correct option: B
x2+b(b2−3c)x+c3=0x^2+b\left(b^2-3 c\right) x+c^3=0
Option B (x2+b(b2−3c)x+c3=0x^2+b(b^2-3 c) x+c^3=0) is correct because it is derived by finding the sum and product of the cubes of the roots of the given quadratic equation.
379 Easy
379 of 528
Application +4 | -1
Value of

x=6+6+6+⋯ up to ∞x=\sqrt{6+\sqrt{6+\sqrt{6+\cdots \text { up to } \infty}}} (1)

is
View answer and explanation
Correct option: A
3
Option A (3) is correct because solving the equation x=6+xx = \sqrt{{6+x}} yields x = 3 as the positive solution.
380 Easy
380 of 528
Analyzing +4 | -1
The number of roots of the quadratic equation 8sec⁡2θ−6sec⁡θ+1=08 \sec ^2 \theta-6 \sec \theta+1=0 is:
View answer and explanation
Correct option: D
0
Option D (0) is correct because the equation 8sec⁡2θ−6sec⁡θ+1=08 \sec^2 \theta - 6 \sec \theta + 1 = 0 has no real roots since the values of sec⁡θ\sec \theta obtained are outside its range.
381 Easy
381 of 528
Application +4 | -1
The equation (cos⁡p−1)x2+cos⁡p  x+sin⁡p=0,\left( {\cos p - 1} \right){x^2} + \cos p\;x + \sin p = 0, in variable xx, has real roots. Then, pp belongs to the interval
View answer and explanation
Correct option: D
(0,π)\left( {0,\pi } \right)
Option D ((0,π)\left( {0,\pi } \right)) is correct because for the quadratic equation to have real roots, the discriminant must be greater than or equal to zero, leading to the condition sin⁡p≥0\sin p \ge 0, which is satisfied in the interval (0,π)\left( {0,\pi } \right).
382 Easy
382 of 528
Application +4 | -1
The roots of the equation (b−c)x2+(c−a)x+(a−b)=0(b-c) x^2+(c-a) x+(a-b)=0 are
View answer and explanation
Correct option: B
a−bb−c,1\frac{a-b}{b-c}, 1
Option B (a−bb−c,1\frac{{a-b}}{{b-c}}, 1) is correct because x=1 is a root, and the product of the roots equals a−bb−c\frac{{a-b}}{{b-c}}.
383 Easy
383 of 528
Application +4 | -1
Let f(x)f\left( x \right) be a quadratic polynomial such that f(−2)+f(3)=0.f\left( { - 2} \right) + f\left( 3 \right) = 0. If one of the roots of f(x)=0f\left( x \right) = 0 is –1, then the sum of the roots of f(x)=0f\left( x \right) = 0 is equal to:
View answer and explanation
Correct option: A
113\frac{{11}}{3}
Option A (113\frac{{11}}{3}) is correct because it is the sum of the roots of the quadratic equation, derived using the given condition f(−2)+f(3)=0f(-2) + f(3) = 0 and the fact that one root is -1.
384 Easy
384 of 528
Analyzing +4 | -1
Number of roots of the equation x−2x−1=1−2x−1x - \frac{2}{{x - 1}} = 1 - \frac{2}{{x - 1}} is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the equation simplifies to x=1, which is not a valid solution due to the original equation's domain restriction.
385 Easy
385 of 528
Application +4 | -1
If (α,β)(\alpha, \beta) are the roots of equation x2−px+q=0x^2-p x+q=0 and (α′,β′)\left(\alpha^{\prime}, \beta^{\prime}\right) are that of x2−p′x+q′=0x^2-p^{\prime} x+q^{\prime}=0, then (α−α′)2+(β−α′)2+(α−β′)2+(β−β′)2=\left(\alpha-\alpha^{\prime}\right)^2+\left(\beta-\alpha^{\prime}\right)^2+\left(\alpha-\beta^{\prime}\right)^2+\left(\beta-\beta^{\prime}\right)^2=
View answer and explanation
Correct option: A
2(p2−2q+p′2−2q′−pp′)2\left(p^2-2 q+p^{\prime 2}-2 q^{\prime}-p p^{\prime}\right)
Option A (2(p2−2q+p′2−2q′−pp′)2\left(p^2-2 q+p^{\prime 2}-2 q^{\prime}-p p^{\prime}\right)) is correct because it is obtained by expanding the given expression and using Vieta's formulas to simplify.
386 Easy
386 of 528
Application +4 | -1
Let f(x)f(x) be a quadratic expression positive for all real xx. If g(x)=f(x)−f′(x)+f′′(x)g(x)=f(x)-f^{\prime}(x)+f^{\prime \prime}(x), then for any real xx
View answer and explanation
Correct option: A
g(x)>0g(x)>0
Option A g(x)>0g(x)>0 is correct because the discriminant of the quadratic expression g(x)g(x) is negative, and the leading coefficient is positive.
387 Easy
387 of 528
Application +4 | -1
Two non-integer roots of the equation

(x2+3x)2−(x2+3x)−6=0\left(x^2+3 x\right)^2-\left(x^2+3 x\right)-6=0 (1)

are⁡\operatorname{are}
View answer and explanation
Correct option: C
12(−3+21),12(−3−21)\frac{1}{2}(-3+\sqrt{21}), \frac{1}{2}(-3-\sqrt{21})
Option C (12(−3+21),12(−3−21)\frac{1}{2}(-3+\sqrt{21}), \frac{1}{2}(-3-\sqrt{21})) is correct because by substituting y=x2+3xy = x^2 + 3x, we can solve the resulting quadratic equation for y, and then solve for x.
388 Easy
388 of 528
Application +4 | -1
If one root of equation x2+px+q=0x^2+p x+q=0 is square of the other then :
View answer and explanation
Correct option: A
p3−q(3p−1)+q2=0p^3-q(3 p-1)+q^2=0
Option A (p3−q(3p−1)+q2=0p^3-q(3 p-1)+q^2=0) is correct because it is derived by relating the roots and coefficients of the quadratic equation and using the condition that one root is the square of the other.
389 Easy
389 of 528
Application +4 | -1
The number of real solutions of x2−2x+2+b−x^2-2 x+2+\mathrm{b}- 11=011=0 is:
View answer and explanation
Correct option: A
0
Option A (0) is correct because the given quadratic equation simplifies to a form that always results in a positive value, indicating no real roots.
390 Easy
390 of 528
Analyzing +4 | -1
The number of real roots of the equation, e4x+e3x−4e2x+ex+1=0{e^{4x}} + {e^{3x}} - 4{e^{2x}} + {e^x} + 1 = 0 is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the equation has only one real root, which is found by transforming the equation and solving for x.
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