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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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361 Hard
361 of 914
Understanding +4 | -1
If roots of ax2+bx+c=0,a,b,c∈R,a≠0a{x^2} + bx + c = 0,a,b,c \in R,a \ne 0 are imaginary then
View answer and explanation
Correct option: A
ac>0ac > 0
Option C (A straight line) is correct because the equation |z - 3| = |z - 5| represents the perpendicular bisector of the line segment joining the points 3 and 5 on the complex plane, which is a straight line.
362 Medium
362 of 914
Application +4 | -1
If a+b+c=0a+b+c=0, then xa2/bc⋅xb2/ca⋅xc2/abx^{a^{2 / b c}} \cdot x^{b^2 / c a} \cdot x^{c^2 / a b} is equal to
View answer and explanation
Correct option: D
x3x^3
The correct answer is D (x3x^3) because when a+b+c=0a+b+c=0, a3+b3+c3=3abca^3+b^3+c^3=3abc, which simplifies the exponent to 3.
363 Medium
363 of 914
Analyzing +4 | -1
Consider two quadratic expressions f(x)=ax2+bx+cf(x)=a x^2+b x+c and g(x)=ax2+px+q(a,b,c,p,q∈R,b≠p)g(x)=a x^2+p x+q(a, b, c, p, q \in R, b \neq p) such that their discriminants are equal. If f(x)=g(x)f(x)=g(x) has a root x=αx=\alpha, then
View answer and explanation
Correct option: A
α\alpha will be AM of the roots of f(x)=0f(x)=0 and g(x)=0g(x)=0
The correct answer is A (α\alpha will be AM of the roots of f(x)=0f(x)=0 and g(x)=0g(x)=0) because it is derived from the given conditions and the properties of quadratic equations.
364 Medium
364 of 914
Application +4 | -1
The product of the roots of 8+x3+8−x3=1\sqrt[3]{8+x}+\sqrt[3]{8-x}=1 is
View answer and explanation
Correct option: A
−189-189
Option A (-189) is correct because it is the product of the solutions obtained after cubing and simplifying the given equation.
365 Easy
365 of 914
Analyzing +4 | -1
If two equations x2+a2=1−2ax{x^2} + {a^2} = 1 - 2ax and x2+b2=1−2bx{x^2} + {b^2} = 1 - 2bx have only one common root, then
View answer and explanation
Correct option: C
a−b=2a - b = 2
Option C (a - b = 2) is correct because by applying the condition for common roots and simplifying the resulting equation, we arrive at a - b = 2.
366 Easy
366 of 914
Analyzing +4 | -1
If the roots of the quadratic equation (4p−p2−5)x2−(2p−1)x+3p=0\left(4 p-p^2-5\right) x^2-(2 p-1) x+3 p=0 lie on either side of unity, the number of integral values of pp is
View answer and explanation
Correct option: B
2
The correct answer is B (2) because the number of integral values of p is found by analyzing the condition where 1 lies between the roots of the quadratic equation.
367 Easy
367 of 914
Application +4 | -1
The α,β\alpha ,\beta are the roots of the equation x2+ax+b=0{x^2} + ax + b = 0, then 1α2+1β2\frac{1}{{{\alpha ^2}}} + \frac{1}{{{\beta ^2}}} is equal to
View answer and explanation
Correct option: A
a2−2bb2\frac{{{a^2} - 2b}}{{{b^2}}}
Option A is correct because it correctly applies the relationship between roots and coefficients of a quadratic equation to find the value of the expression.
368 Easy
368 of 914
Application +4 | -1
If 32+72i\frac{3}{2} + \frac{7}{2}i is a solution of the equation ax2−6x+b=0a{x^2} - 6x + b = 0, where aa and bb are real numbers, then the value of a+ba + b is equal to
View answer and explanation
Correct option: D
31
Option D (31) is correct because finding the conjugate root and using Vieta's formulas leads to a = 2 and b = 29, so a + b = 31.
369 Easy
369 of 914
Application +4 | -1
The quadratic equation in xx such that the arithmetic mean of its roots is 5 and geometric mean of the roots is 4, is given by
View answer and explanation
Correct option: B
x2−10x+16=0{x^2} - 10x + 16 = 0
Option B (x2−10x+16=0{x^2} - 10x + 16 = 0) is correct because it is formed using the given arithmetic and geometric means of the roots.
370 Easy
370 of 914
Application +4 | -1
If sin⁡α,cos⁡α\sin \alpha ,\cos \alpha are the roots of the equation ax2+bx+c=0,a{x^2} + bx + c = 0, then
View answer and explanation
Correct option: A
a2−b2+2ac=0{a^2} - {b^2} + 2ac = 0
Option D (a2−b2+2ac=0{a^2} - {b^2} + 2ac = 0) is correct because by using the relationships between the roots and coefficients of the quadratic equation and the trigonometric identity sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1, we can derive the given relation.
371 Medium
371 of 914
Application +4 | -1
If α\alpha and β\beta are roots of the equation x2+px+3p4=0x^{2}+p x+\frac{3 p}{4}=0, such that ∣α−β∣=10|\alpha-\beta|=\sqrt{10}, then pp belongs to the set :
View answer and explanation
Correct option: C
{−2,5}\{-2,5\}
Option C ({−2,5}\{-2,5\}) is correct because solving for p using the given condition ∣α−β∣=10|\alpha-\beta|=\sqrt{10} yields the set {−2,5}\{-2,5\}.
372 Easy
372 of 914
Application +4 | -1
If one root of equation x2+ax+12=0{x^2} + ax + 12 = 0 is 4 while the equation x2+ax+b=0{x^2} + ax + b = 0 has equal roots, then the value of bb is
View answer and explanation
Correct option: B
494\frac{{49}}{4}
Option B (494\frac{{49}}{4}) is correct because the value of b is calculated using the condition for equal roots of a quadratic equation.
373 Easy
373 of 914
Application +4 | -1
Let α\alpha and β\beta be the roots of px2+qx+r=0p x^2+q x+r=0, p≠0p \neq 0. If p,q,rp, q, r are in A.P. and 1α+1β=4\frac{1}{\alpha}+\frac{1}{\beta}=4, then value of ∣α−β∣|\alpha-\beta| is
View answer and explanation
Correct option: D
2913\frac{2}{9} \sqrt{13}
Option D (2913\frac{2}{9} \sqrt{13}) is correct because using the given conditions that p, q, r are in A.P. and the relationship between the roots, we can find the value of |α−β\alpha - \beta|.
374 Easy
374 of 914
Application +4 | -1
If the roots of the equation 1x+p+1x+q=1r\frac{1}{{x + p}} + \frac{1}{{x + q}} = \frac{1}{r} are equal in magnitude but opposite in sign, then the product of the roots will be
View answer and explanation
Correct option: B
−(p2+q2)2- \frac{{\left( {{p^2} + {q^2}} \right)}}{2}
Option B (−(p2+q2)2- \frac{{{\left( {{{{p^2} + {q^2}}}} \right)}}}{{{{2}}}}) is correct because it is the product of the roots calculated using the given condition that the roots are equal in magnitude but opposite in sign.
375 Easy
375 of 914
Analyzing +4 | -1
If 1α\frac{1}{\sqrt{\alpha}} and 1β\frac{1}{\sqrt{\beta}} are the roots of the equation, ax2+bx+1=0(a≠0,a,b,∈R)a x^{2}+b x+1=0(a \neq 0, a, b, \in R), then the equation, x(x+b3)+(a3−3abx)=0x\left(x+b^{3}\right)+\left(a^{3}-3 a b x\right)=0 as roots :
View answer and explanation
Correct option: A
α3/2\alpha^{3 / 2} and β3/2\beta^{3 / 2}
Option A (α3/2\alpha^{3 / 2} and β3/2\beta^{3 / 2}) is correct because by using the relationships between roots and coefficients, we can find the roots of the new equation.

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376 Easy
376 of 914
Analyzing +4 | -1
Let p,qp, q and rr be real numbers (p≠q,r≠0)(\mathrm{p} \neq \mathrm{q}, \mathrm{r} \neq 0), such that the roots of the equation 1x+p+1x+q=1r\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r} are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to.
View answer and explanation
Correct option: B
p2+q2p^{2}+q^{2}
Option B (p2+q2p^{2}+q^{2}) is correct because by setting the sum of the roots to zero (since they are equal in magnitude but opposite in sign) and using the quadratic formula, we can find the sum of the squares of the roots.
377 Easy
377 of 914
Analyzing +4 | -1
If α\alpha and β\beta are the roots of the quadratic equation, x2+xsin⁡θ−2sin⁡θ=0,θ∈(0,π2){x^2} + x\sin \theta - 2\sin \theta = 0,\theta \in \left( {0,\frac{\pi }{2}} \right) then α12+β12(α−12+β−12)(α−β)24\frac{{{\alpha ^{12}} + {\beta ^{12}}}}{{\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}} is equal to
View answer and explanation
Correct option: A
212(sin⁡θ+8)12\frac{{{2^{12}}}}{{{{\left( {\sin \theta + 8} \right)}^{12}}}}
Option A (212(sin⁡θ+8)12\frac{{{2^{12}}}}{{{{\left( {\sin \theta + 8} \right)}^{12}}}}) is correct because it is the simplified form of the given expression using the relationship between the roots and coefficients of the quadratic equation.
378 Medium
378 of 914
Application +4 | -1
If x2−3x+2x^2-3 x+2 is one of the factors of the expression x4−px2+qx^4-p x^2+q, then
View answer and explanation
Correct option: B
p=5p=5 and q=4q=4
Option B (p=5 and q=4) is correct because when x4−px2+qx^4-px^2+q is divided by x2−3x+2x^2-3x+2, the remainder must be zero, leading to p=5 and q=4.
379 Easy
379 of 914
Analyzing +4 | -1
If the equation x2+px+q=0{x^2} + px + q = 0 has roots uu and υυ where p,qp,q are non-zero constants. Then,
View answer and explanation
Correct option: A
qx2+px+1=0q{x^2} + px + 1 = 0 has roots 1u\frac{1}{u} and 1v\frac{1}{v}
Option A (qx2+px+1=0q{x^2} + px + 1 = 0 has roots 1u\frac{1}{u} and 1v\frac{1}{v}) is correct because if uu and vv are roots of x2+px+q=0{x^2} + px + q = 0, then the equation with roots 1u\frac{1}{u} and 1v\frac{1}{v} is obtained by substituting x=1yx = \frac{1}{y} in the original equation.
380 Medium
380 of 914
Analyzing +4 | -1
If a,b,c,d\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} and p\mathrm{p} are distinct non zero real numbers such that (a2+b2+c2)p2−2(ab+bc+cd)p\left(a^2+b^2+c^2\right) p^2-2(a b+b c+c d) p +(b2+c2+d2)≤0+\left(b^2+c^2+d^2\right) \leq 0 then a,b,c,da, b, c, d are in
View answer and explanation
Correct option: B
G.P.
Option B (G.P.) is correct because the given condition leads to the ratios of consecutive terms being equal, indicating a geometric progression.
381 Medium
381 of 914
Analyzing +4 | -1
The set of all real numbers satisfying the inequation x(log⁡10x)2−3(log⁡10x)+1>1000,{x^{{{\left( {{{\log }_{10}}x} \right)}^2} - 3\left( {{{\log }_{10}}x} \right) + 1}} > 1000, is
View answer and explanation
Correct option: D
None of these
The correct answer is None of these. After solving the inequality using logarithms, the set of real numbers satisfying the inequation is (0,1) U (1000, ∞).
382 Medium
382 of 914
Application +4 | -1
If α\alpha and β\beta are the roots of equation ax2+bx+c=0a x^2+b x+c=0, then the sum of the roots of the equation a2x2+(b2−2ac)x+b2−4ac=0a^2 x^2+\left(b^2-2 a c\right) x+b^2-4 a c=0 is given by -
View answer and explanation
Correct option: D
−(α2+β2)-\left(\alpha^2+\beta^2\right)
Option D (−(α2+β2)-\left(\alpha^2+\beta^2\right)) is correct because it represents the sum of the roots of the transformed quadratic equation in terms of the original roots.
383 Easy
383 of 914
Understanding +4 | -1
In the equation a4x4+a3x3+a2x2+a1x+a0=0{a_4}{x^4} + {a_3}{x^3} + {a_2}{x^2} + {a_1}x + {a_0} = 0 roots of the equation are αi,    i=1,  2,  3,  4.{\alpha _i},\;\;i = 1,\;2,\;3,\;4. Now, xx is replaced by x−1x - 1, now roots of new equation are
View answer and explanation
Correct option: B
αi+1,  i=1,  2,  3,  4{\alpha _i} + 1,\;i = 1,\;2,\;3,\;4
Option B is correct because replacing x with x-1 in the polynomial equation results in roots that are the original roots plus 1.
384 Medium
384 of 914
Application +4 | -1
The 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 sign, are
View answer and explanation
Correct option: A
1<a<21 < a < 2
Option A (1<a<21 < a < 2) is correct because the roots have opposite signs when the product of the roots is negative, leading to the inequality (a−1)(a−2)<0(a-1)(a-2) < 0.
385 Easy
385 of 914
Analyzing +4 | -1
If x2−3x+2x^2-3 x+2 be a factor of x4−px2+qx^4-p x^2+q, then (p,q)=(p, q)= ?
View answer and explanation
Correct option: D
(5,4)(5,4)
Option D ((5,4)) is correct because substituting the roots of x2−3x+2=0x^2-3x+2=0 into x4−px2+q=0x^4-px^2+q=0 and solving the resulting system of equations yields p=5p=5 and q=4q=4.
386 Medium
386 of 914
Application +4 | -1
If the first, second and the last terms of an A.P. are a,b,ca, b, c respectively, then the sum is
(a+b)(a+c−2b)2(b−a)\frac{(a+b)(a+c-2 b)}{2(b-a)}
(b+c)(a+b−2c)2(b−a)\frac{(b+c)(a+b-2 c)}{2(b-a)}
View answer and explanation
Correct option: C
(b+c)(a+b−2c)2(b−a)\frac{{\left( {b + c} \right)\left( {a + b - 2c} \right)}}{{2\left( {b - a} \right)}}
Option C ((b+c)(a+b−2c)2(b−a)\frac{{\left( {b + c} \right)\left( {a + b - 2c} \right)}}{{2\left( {b - a} \right)}}) is correct because it represents the sum of an arithmetic progression with given first, second, and last terms.
387 Easy
387 of 914
Application +4 | -1
If α+β=−2\alpha + \beta = - 2 and α3+β3=−56{\alpha ^3} + {\beta ^3} = - 56, then the quadratic equation whose roots are α\alpha and β\beta is
View answer and explanation
Correct option: D
x2+2x−8=0{x^2} + 2x - 8 = 0
Option D (x2+2x−8=0{x^2} + 2x - 8 = 0) is correct because given α+β=−2\alpha + \beta = - 2 and α3+β3=−56{\alpha ^3} + {\beta ^3} = - 56, we can find αβ=−8\alpha\beta = -8 and thus form the quadratic equation.
388 Easy
388 of 914
Application +4 | -1
If domain of f(x)=x2+bx+4f(x)=\sqrt{x^2+b x+4} is R\mathbf{R}, then maximum possible integral value of bb is
View answer and explanation
Correct option: C
4
Option C (4) is correct because it is the maximum integer value of b for which the discriminant of the quadratic inside the square root is non-positive, ensuring the expression is defined for all real numbers.
389 Easy
389 of 914
Analyzing +4 | -1
The total number of roots of the equation ∣x−x2−1∣=∣2x−3−x2∣\left| {x - {x^2} - 1} \right| = \left| {2x - 3 - {x^2}} \right| is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the equation simplifies to a linear equation with only one real root.
390 Medium
390 of 914
Application +4 | -1
If α,β\alpha, \beta are the roots of quadratic equation x2+px+q=0x^2+p x+q=0 and γ,δ\gamma, \delta are the roots of x2+px−r=x^2+p x-r= 0 , then (α−γ)⋅(α−δ)(\alpha-\gamma) \cdot(\alpha-\delta) is equal to :
View answer and explanation
Correct option: C
−(q+r)-(q+r)
Option C (−(q+r)-(q+r)) is correct because by substituting α\alpha into the second equation and using the relationships between roots and coefficients, the expression simplifies to −(q+r)-(q+r).
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