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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Easy

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451 Easy
451 of 528
Application +4 | -1
If the roots of the equation 3x2−6x+5=03{x^2} - 6x + 5 = 0 are α\alpha and β\beta, then the equation whose roots are α+β\alpha + \beta and 2α+β\frac{2}{{\alpha + \beta }} will be
View answer and explanation
Correct option: D
x2−3x+2=0{x^2} - 3x + 2 = 0
Option D (x2−3x+2=0{x^2} - 3x + 2 = 0) is correct because it is the quadratic equation formed with roots α+β\alpha + \beta and 2α+β\frac{2}{{\alpha + \beta}}.
452 Easy
452 of 528
Application +4 | -1
If 2+i32 + i\sqrt 3 is a root of the equation x2+px+q=0,{x^2} + px + q = 0, where pp and qq are real, then (p,q)\left( {p,q} \right) is equal to
View answer and explanation
Correct option: A
(−4,  7)\left( { - 4,\;7} \right)
Option A ((−4,  7)\left( { - 4,\;7} \right)) is correct because if 2+i32 + i\sqrt 3 is a root, then its conjugate is also a root, and we can find p and q using the sum and product of roots.
453 Easy
453 of 528
Application +4 | -1
The number of solutions of

4−x+x+9=5\sqrt{4-x}+\sqrt{x+9}=5 (1)

is:
View answer and explanation
Correct option: C
2
Option C (2) is correct because after solving the equation, we find two solutions, x=0 and x=-5, both of which satisfy the original equation.
454 Easy
454 of 528
Application +4 | -1
If y=2[x]+3y=2[x]+3 and y=3[x−2]y=3[x-2], then [x+y]=?[x+y]=?
View answer and explanation
Correct option: C
30
Option C (30) is correct because solving the given equations for [x] and y, we find [x] = 9 and y = 21, so [x+y] = [9+21] = 30.
455 Easy
455 of 528
Analyzing +4 | -1
The number of solutions of 3∣x∣=∣2−∣x∣∣{3^{\left| x \right|}} = \left| {2 - \left| x \right|} \right| is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the graphs of the exponential and absolute value functions intersect at two points.
456 Easy
456 of 528
Application +4 | -1
If the roots of the quadratic equation x2+px+q=0x^2+p x+q=0 are tan⁡30∘\tan 30^{\circ} and tan⁡15∘\tan 15^{\circ}, respectively, the value of 2+q−p2+q-p is
View answer and explanation
Correct option: B
3
Option B (3) is correct because using the tangent addition formula, we can relate the sum and product of the roots to find the value of 2+q−p2+q-p.
457 Easy
457 of 528
Analyzing +4 | -1
if x2−bxax−c=λ−1λ+1\frac{{{x^2} - bx}}{{ax - c}} = \frac{{\lambda - 1}}{{\lambda + 1}} has roots equal in magnitude and opposite in sign then the value of λ\lambda is
View answer and explanation
Correct option: A
a−ba+b\frac{{a - b}}{{a + b}}
Option A is correct because the sum of the roots being zero leads to the given expression for lambda.
458 Easy
458 of 528
Application +4 | -1
Product of roots of the equation 13−x2=x+5\sqrt{13-x^2}=x+5 is
View answer and explanation
Correct option: C
6
Option C (6) is correct because the product of the roots of the equation is 6.
459 Easy
459 of 528
Analyzing +4 | -1
The number of real solutions of the equation (910)x=−3+x−x2{\left( {\frac{9}{{10}}} \right)^x} = - 3 + x - {x^2} is
View answer and explanation
Correct option: A
None
Option A (None) is correct because the left-hand side (LHS) is always positive, while the right-hand side (RHS) is always negative, so there are no real solutions.
460 Easy
460 of 528
Application +4 | -1
Let α\alpha and β\beta be two real roots of the equation(k+1)tan⁡2x−2⋅λtan⁡x=(1−k),(k + 1){\tan ^2}x - \sqrt 2 \cdot \mathrm{ }\lambda \tan x = (1 - k), where k(≠−1)k( \ne - 1) and λ\mathrm{ }\lambda are real numbers. If tan⁡2(α+β)=50{\tan ^2}(\alpha \mathrm{ } + \mathrm{ }\beta ) = 50 , then a value of λ\lambda is
View answer and explanation
Correct option: B
10
Option B (10) is correct because it is one of the values of λ\lambda obtained by solving the trigonometric equation using the given condition tan⁡2(α+β)=50{\tan ^2}(\alpha + \mathrm{ }\beta ) = 50.
461 Easy
461 of 528
Application +4 | -1
The values of aa which make the expression x2−ax+x^2-a x+ 1−2a21-2 a^2 always positive for real values of xx, are
View answer and explanation
Correct option: A
−23<a<23-\frac{2}{3} < a < \frac{2}{3}
Option A (-2/3 < a < 2/3) is correct because it ensures the discriminant of the quadratic is negative, making the expression always positive.
462 Easy
462 of 528
Analyzing +4 | -1
If a,b,ca,b,c are positive real numbers such that a+b+c=2a + b + c = 2 then, which one of the following is true?
View answer and explanation
Correct option: A
(2−a)(2−b)(2−c)≥8abc\left( {2 - a} \right)\left( {2 - b} \right)\left( {2 - c} \right) \ge 8abc
Option A ((2−a)(2−b)(2−c)≥8abc\left( {2 - a} \right)\left( {2 - b} \right)\left( {2 - c} \right) \ge 8abc) is correct because it is derived from applying the AM-GM inequality to the given conditions.
463 Easy
463 of 528
Analyzing +4 | -1
If (1+k)tan⁡2x−4tan⁡x−1+k=0\left( {1 + k} \right){\tan ^2}x - 4\tan x - 1 + k = 0 has real roots tan⁡x1\tan {x_1} and tan⁡x2,\tan {x_2}, then
View answer and explanation
Correct option: A
k2≤5{k^2} \le 5
Option A (k2≤5{k^2} \le 5) is correct because it represents the condition for the quadratic equation to have real roots, based on the discriminant.
464 Easy
464 of 528
Application +4 | -1
Suppose p∈Rp \in \mathbf{R} and α,β\alpha, \beta are roots of x2−px+12p2=0x^2-p x+\frac{1}{2 p^2}=0, then the minimum value of α4+β4\alpha^4+\beta^4 is
View answer and explanation
Correct option: C
2−2\sqrt{2}-2
Option C (2−2\sqrt{2}-2) is correct because by using AM-GM inequality, we can find the minimum value of the expression, which occurs when p4=12p4p^4 = \frac{1}{2p^4}.
465 Easy
465 of 528
Analyzing +4 | -1
The minimum value of the sum of the squares of the roots of x2+(3−a)x+1=2ax^2+(3-a) x+1=2 a is
View answer and explanation
Correct option: C
6
Option C (6) is correct because the minimum value of the sum of squares of the roots is found by completing the square and finding the vertex of the resulting quadratic.

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466 Easy
466 of 528
Analyzing +4 | -1
If x∈R,x \in R, the least value of the expression x2−6x+5x2+2x+1,\frac{{{x^2} - 6x + 5}}{{{x^2} + 2x + 1}}, is
View answer and explanation
Correct option: C
−13\frac{{ - 1}}{3}
Option C (-1/3) is correct because the minimum value of the expression is found by setting the discriminant greater than or equal to zero and solving for y.
467 Easy
467 of 528
Analyzing +4 | -1
The roots of the quadratic equation
(a+b−2c)x2−(2a−b−c)x+(a−2b+c)=0\left( {a + b - 2c} \right){x^2} - \left( {2a - b - c} \right)x + \left( {a - 2b + c} \right) = 0 are
View answer and explanation
Correct option: D
None of these
Option B (At least two real roots) is correct because the product of the discriminants of the two quadratic factors is always non-negative, implying at least one factor has real roots.
468 Easy
468 of 528
Application +4 | -1
If α,β\alpha, \beta are the roots of the quadratic equation ax2+bx+c=0a x^2+b x+c=0, then the quadratic equation whose roots are α3,β3\alpha^3, \beta^3 is
View answer and explanation
Correct option: A
a3y2+(b3−3abc)y+c3=0a^3 y^2+\left(b^3-3 a b c\right) y+c^3=0
Option A (a3y2+(b3−3abc)y+c3=0a^3 y^2+\left(b^3-3 a b c\right) y+c^3=0) is correct because it is the transformed quadratic equation with roots α3\alpha^3 and β3\beta^3.
469 Easy
469 of 528
Application +4 | -1
If x2r2−r−6+y2r2−6r+5=1\frac{x^2}{r^2-r-6}+\frac{y^2}{r^2-6 r+5}=1 will represents the ellipse, if rr lies in the interval
View answer and explanation
Correct option: A
(−∞,−2)∪(5,∞)(-\infty,-2) \cup(5, \infty)
The correct answer is A ((−∞,−2)∪(5,∞))((-\infty,-2) \cup(5, \infty)). For the equation to represent an ellipse, both denominators must be positive.
470 Easy
470 of 528
Application +4 | -1
If α,β\alpha ,\beta are the roots of ax2+bx+c=0a{x^2} + bx + c = 0 the equation whose roots are 2+α,2+β2 + \alpha ,2 + \beta is
View answer and explanation
Correct option: D
ax2+x(b−4  a)+4  a−2  b+c=0a{x^2} + x\left( {b - 4\;a} \right) + 4\;a - 2\;b + c = 0
Option D (ax2+x(b−4  a)+4  a−2  b+c=0a{x^2} + x\left( {b - 4\;a} \right) + 4\;a - 2\;b + c = 0) is correct because by transforming the original equation using the relationship between the roots, we obtain the new equation with roots 2 + α\alpha and 2 + β\beta.
471 Easy
471 of 528
Application +4 | -1
Sum of the roots of the equation is

4x−3(2x+3)+128=04^x-3\left(2^{x+3}\right)+128=0 (1)
is
View answer and explanation
Correct option: C
7
Option C (7) is correct because the sum of the roots is obtained by substituting y=2xy = 2^x and solving the resulting quadratic equation for x.
472 Easy
472 of 528
Application +4 | -1
If aa is the minimum root of the equation x2−3∣x∣−2=0x^2-3|x|-2=0, then value of −1/a-1 / a is
View answer and explanation
Correct option: A
(17−3)/4(\sqrt{17}-3) / 4
Option A ((17−3)/4(\sqrt{{17}}-3) / 4) is correct because it is the negative reciprocal of the minimum root of the given equation after solving for |x| and considering both positive and negative roots.
473 Easy
473 of 528
Application +4 | -1
If the product 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 2, then the sum of roots is
View answer and explanation
Correct option: B
−1- 1
Option B (-1) is correct because by finding 'a' using the product of roots condition and substituting it into the sum of roots formula, we get -1.
474 Easy
474 of 528
Application +4 | -1
The values of aa and bb so that x4+12x3+46x2+ax+bx^{4}+12 x^{3}+46 x^{2}+a x+b is square of quadratic expression are, respectively,
View answer and explanation
Correct option: A
60,25
Option A (60, 25) is correct because when a=60a = 60 and b=25b = 25, the given expression becomes a perfect square of a quadratic expression.
475 Easy
475 of 528
Analyzing +4 | -1
If x∈R,x \in R, then the expression 9x−3x+1{9^x} - {3^x} + 1 assumes
View answer and explanation
Correct option: C
All real values greater than 34\frac{3}{4}
Option C (All real values greater than 34\frac{3}{4}) is correct because completing the square shows the minimum value of the expression is 34\frac{3}{4}.
476 Easy
476 of 528
Analyzing +4 | -1
The value of pp for which both the roots of the equation 4  x2−20  px+(25  p2+15  p−66)=04\;{x^2} - 20\;px + \left( {25\;{p^2} + 15\;p - 66} \right) = 0 are less than 2, lies in
View answer and explanation
Correct option: D
(−∞,−1)\left( { - \infty , - 1} \right)
Option D ((−∞,−1)\left( { - \infty , - 1} \right)) is correct because it satisfies the conditions for both roots to be less than 2.
477 Easy
477 of 528
Application +4 | -1
If the equations ax2+bc+c=0a{x^2} + bc + c = 0 and 2x2+3x+4=02{x^2} + 3x + 4 = 0 have a common root, then a  :b  :ca\;:b\;:c
View answer and explanation
Correct option: A
2  :3  :42\;:3\;:4
The correct answer is A (2  :  3  :  42\;:\;3\;:\;4) because if two quadratic equations have a common root and one has complex roots, they must have both roots in common, implying their coefficients are proportional.
478 Easy
478 of 528
Analyzing +4 | -1
If the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0 of the form k+1k   and   k+2k+1,   then   (a+b+c)2\frac{{k + 1}}{k}\;\,\mathrm{and}\,\;\frac{{k + 2}}{{k + 1}},\;\,\mathrm{then}\,\;{\left( {a + b + c} \right)^2} is equal to
View answer and explanation
Correct option: C
b2−4ac{b^2} - 4ac
The correct answer is C (b2−4ac{b^2} - 4ac) because by using the given roots and Vieta's formulas, we can derive the relationship and simplify to find the value of (a+b+c)2(a+b+c)^2.
479 Easy
479 of 528
Analyzing +4 | -1
If α,β\alpha ,\beta are ther roots of ax2+bx+c=0a{x^2} + bx + c = 0 and α+β,α2+β2,α3+β3\alpha + \beta ,{\alpha ^2} + {\beta ^2},{\alpha ^3} + {\beta ^3} are in G.P., where Δ=−b2−4ac\Delta = - {b^2} - 4ac, then
View answer and explanation
Correct option: D
cΔ=0c\Delta = 0
Option D (cΔ=0c\Delta = 0) is correct because the given condition that α+β,α2+β2,α3+β3\alpha + \beta, \alpha^2 + \beta^2, \alpha^3 + \beta^3 are in G.P. leads to the equation ac(b2−4ac)=0ac(b^2 - 4ac) = 0, which simplifies to cΔ=0c\Delta = 0.
480 Easy
480 of 528
Application +4 | -1
The graph of the function y=16x2+8(a+5)x−7  a−5y = 16{x^2} + 8\left( {a + 5} \right)x - 7\;a - 5 is strictly above the xx-axis, then 'a' must satisfy the inequality
View answer and explanation
Correct option: A
−15<a<−2- 15 < a < - 2
Option A (-15 < a < -2) is correct because for the quadratic to be strictly above the x-axis, its discriminant must be negative, leading to the specified inequality for 'a'.
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