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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Easy

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Showing 61-90 of 528 questions
61 Easy
61 of 528
Analyzing +4 | -1
Solution set of the equation 32x2−2⋅3x2+x+6+32(x+6)=03^{2 x^2}-2 \cdot 3^{x^2+x+6}+3^{2(x+6)}=0 is
View answer and explanation
Correct option: C
{−2,3}\{-2,3\}
The correct answer is C ({-2,3}) because it is the solution set obtained by simplifying the given exponential equation and solving the resulting quadratic equation.
62 Easy
62 of 528
Application +4 | -1
Let α,β\alpha, \beta be roots of ax2+bx+c=0a x^2+b x+c=0, where ac≠0a c \neq 0. Roots of cx2−bx+a=0c x^2-b x+a=0 are
View answer and explanation
Correct option: B
−1/α,−1/β-1 / \alpha,-1 / \beta
Option A (1/α,1/β1 / \alpha, 1 / \beta) is incorrect. The correct answer is that the roots of cx2−bx+a=0cx^2 - bx + a = 0 are 1/α1/\alpha and 1/β1/\beta.
63 Easy
63 of 528
Analyzing +4 | -1
The value of xx which satisfy the equation (5x2−8x+3)−(5x2−9x+4)=(2x2−2x)−(2x2−3x+1)\sqrt {\left( {5{x^2} - 8x + 3} \right)} - \sqrt {\left( {5{x^2} - 9x + 4} \right)} = \sqrt {\left( {2{x^2} - 2x} \right)} - \sqrt {\left( {2{x^2} - 3x + 1} \right)} is
View answer and explanation
Correct option: C
1
Option C (1) is correct because by factoring and simplifying the equation, we find that x=1 is the only solution.
64 Easy
64 of 528
Analyzing +4 | -1
Let α,β\alpha, \beta be the roots of the equation x2−2x+6=0x^2-\sqrt{2} x+\sqrt{6}=0 and 1α2+1,1β2+1,1β2+1\frac{1}{\alpha^2}+1, \frac{1}{\beta^2}+1, \frac{1}{\beta^2}+1 be the roots of the equation x2+ax+b=0x^2+a x+b=0. Then the roots of the equation x2−(a+b−2)x+(a+b+2)x^2-(a+b-2) x+(a+b+2) =0=0 are
View answer and explanation
Correct option: B
real and both negative
Option B (real and both negative) is correct because the roots of the quadratic equation 6x2+17x+7=06x^2 + 17x + 7 = 0 are real and both negative.
65 Easy
65 of 528
Analyzing +4 | -1
The solution set of x2+2≤3x≤2x2−5,{x^2} + 2 \le 3x \le 2{x^2} - 5, is
View answer and explanation
Correct option: A
Φ\mathrm{\Phi }
Option A (Φ\Phi) is correct because there is no value of x that satisfies both inequalities x2+2≤3x{x^2} + 2 \le 3x and 3x≤2x2−53x \le 2{x^2} - 5 simultaneously.
66 Easy
66 of 528
Application +4 | -1
If the roots of the equation x2−bx+c=0{x^2} - bx + c = 0 are two consecutive integers, then b2−4c{b^2} - 4c is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the difference between the roots is used to derive the value of the expression b2 - 4c.
67 Easy
67 of 528
Application +4 | -1
The number of solutions of

x+1−x−1=1(x∈R)\sqrt{x+1}-\sqrt{x-1}=1 (x \in \mathbf{R}) (1)

is
View answer and explanation
Correct option: A
1
Option A (1) is correct because solving the equation x+1−x−1=1\sqrt{{x+1}}-\sqrt{{x-1}}=1 leads to a single valid solution, x=5/4x = 5/4.
68 Easy
68 of 528
Application +4 | -1
If the difference between the roots of the equation x2+ax+1=0x^2+a x+1=0 is less than 5\sqrt{5}, the set of possible values of aa, is
View answer and explanation
Correct option: A
(−3,3)(-3,3)
Option A ((-3,3)) is correct because the difference between the roots being less than 5\sqrt{5} leads to the inequality a2<9a^2 < 9, which gives the interval (-3,3).
69 Easy
69 of 528
Application +4 | -1
If 2sin⁡2π82{\sin ^2}\frac{\pi }{8} is a root of the equation x2+ax+b=0{x^2} + ax + b = 0, where aa and bb are rational numbers, then a−ba - b is equal to
View answer and explanation
Correct option: A
−52- \frac{5}{2}
Option A (−52\frac{{ - 5}}{2}) is correct because it is the value of a - b, where a and b are coefficients of the quadratic equation and 2sin⁡2π82{\sin ^2}\frac{\pi }{8} is one of its roots.
70 Easy
70 of 528
Application +4 | -1
If a,b,c∈Ra,b,c \in R and a+b+c=0,a + b + c = 0, then the quadratic equation 4ax2+3bx+2c=04a{x^2} + 3bx + 2c = 0 has
View answer and explanation
Correct option: C
Real roots
Option C (Real roots) is correct because the discriminant of the quadratic equation is always positive, ensuring real roots.
71 Easy
71 of 528
Application +4 | -1
If the roots of the equation x2+a2=8x+6a{x^2} + {a^2} = 8x + 6a are real, then aa belongs to the interval
View answer and explanation
Correct option: B
[−2,  8]\left[ { - 2,\;8} \right]
Option B ([ - 2, 8 ]) is correct because the discriminant of the quadratic equation must be greater than or equal to zero for real roots, leading to the interval [ -2, 8 ].
72 Easy
72 of 528
Application +4 | -1
If x∈Rx \in \mathbf{R}, the number of solutions of 2x+1−\sqrt{2 x+1}- 2x−1=1\sqrt{2 x-1}=1 is
View answer and explanation
Correct option: B
1
Option B (1) is correct because after squaring and simplifying, only one solution satisfies the original equation's domain.
73 Easy
73 of 528
Application +4 | -1
Let x2+x+1x^{2}+x+1 be divisible by 3 . If xx is divided by 3 , the remainder will be
View answer and explanation
Correct option: B
1
Option B (1) is correct because when x is divided by 3, the remainder is 1, satisfying the given condition that x2+x+1x^2 + x + 1 is divisible by 3.
74 Easy
74 of 528
Analyzing +4 | -1
A value of bb for which the equations x2+bx−1=0x2+x+b=0{x^2} + bx - 1 = 0\quad {x^2} + x + b = 0 have one root in common, is
View answer and explanation
Correct option: B
−i3,i=−1-i \sqrt{3}, i=\sqrt{-1}
Option B (−i3,i=−1-i \sqrt{3}, i=\sqrt{-1}) is correct because it is a value of bb for which the two given equations have a common root.
75 Easy
75 of 528
Analyzing +4 | -1
Let α\alpha and β\beta be the roots of the equation x2+x^2+ (2i−1)=0(2 i-1)=0. Then, the value of ∣α8+β8∣\left|\alpha^8+\beta^8\right| is equal to:
View answer and explanation
Correct option: A
50
Option A (50) is correct because, with α2=β2=1−2i\alpha^2 = \beta^2 = 1 - 2i, we find α8=β8=−7+24i\alpha^8 = \beta^8 = -7 + 24i, and then ∣α8+β8∣=2∣−7+24i∣=50|\alpha^8 + \beta^8| = 2|-7 + 24i| = 50.

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76 Easy
76 of 528
Application +4 | -1
In the equation x2+px+q=0x^2+p x+q=0, the coefficient of xx was incorrectly written as 17 instead of 13 , then the roots were found to be -2 and -15 . The correct equation is
View answer and explanation
Correct option: C
x2+13x+30=0x^2+13 x+30=0
Option C (x2+13x+30=0x^2+13 x+30=0) is correct because the constant term was read correctly, and the correct coefficient of x is 13.
77 Easy
77 of 528
Analyzing +4 | -1
The quadratic equations
x2−6x+a=0{x^2} - 6x + a = 0
And x2−cx+6=0{x^2} - cx + 6 = 0
Have one root in common. The other roots of the first and second equations are integers in the ratio 4:34:3. Then the common root is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the common root is found by setting up equations based on the given information about the roots of the two quadratic equations and solving for the common root.
78 Easy
78 of 528
Application +4 | -1
The greatest intergral value of mm for which the equation (2m−1)x2−4x+(2m−1)(2 m-1) x^2-4 x+(2 m-1) has real roots is given by
View answer and explanation
Correct option: B
1
Option B (1) is correct because the discriminant must be greater than or equal to zero for real roots, leading to the inequality −12<m<32-\frac{1}{2} < m < \frac{3}{2}, and the greatest integer value of m is 1.
79 Easy
79 of 528
Application +4 | -1
Suppose α,β\alpha, \beta are roots of x2−7x+8=0x^2-7 x+8=0, with α\alpha >β>\beta, then value of 16α+3β2−19β\frac{16}{\alpha}+3 \beta^2-19 \beta is
View answer and explanation
Correct option: A
-10
Option A (-10) is correct because after simplification and substitution, the value of the expression is -10.
80 Easy
80 of 528
Application +4 | -1
If α,β\alpha ,\beta are the roots of the equation x2−(1+n2)x+12(1+n2+n4)=0{x^2} - \left( {1 + {n^2}} \right)x + \frac{1}{2}\left( {1 + {n^2} + {n^4}} \right) = 0, then α2+β2{\alpha ^2} + {\beta ^2} is
View answer and explanation
Correct option: A
n2{n^2}
Option A (n2{n^2}) is correct because the expression α2+β2{\alpha ^2} + {\beta ^2} is simplified using the sum and product of the roots of the quadratic equation.
81 Easy
81 of 528
Analyzing +4 | -1
A quadratic equation the product of whose roots x1x_1 and x2x_2 is equal to 4 and satisfying the relation x1x1−1+x2x2−1=2\frac{x_1}{x_1-1}+\frac{x_2}{x_2-1}=2, is
View answer and explanation
Correct option: A
x2−2x+4=0x^2-2 x+4=0
The correct answer is A (x2−2x+4=0x^2-2 x+4=0) because it satisfies both the product of roots condition and the given relation between the roots.
82 Easy
82 of 528
Application +4 | -1
The condition f(x)=x3+px2+qx+r(x∈R)f\left( x \right) = {x^3} + p{x^2} + qx + r\left( {x \in R} \right) to have no extreme value, is
View answer and explanation
Correct option: A
p2<3q{p^2} < 3q
Option A (p2<3q{p^2} < 3q) is correct because it provides the condition for the given cubic function to have no extreme values, which occurs when its derivative has no real roots.
83 Easy
83 of 528
Application +4 | -1
If α,β\alpha ,\beta are roots of the equation 6x2−5x+1=0,6{x^2} - 5x + 1 = 0, then the value of tan⁡−1α+tan⁡−1β{\tan ^{ - 1}}\alpha + {\tan ^{ - 1}}\beta is
View answer and explanation
Correct option: B
π4\frac{\pi }{4}
Option B (π4\frac{\pi}{4}) is correct because by using the formula for the sum of two arctangents and substituting the sum and product of the roots, we obtain tan⁡−1(1)=π4\tan^{-1}(1) = \frac{\pi}{4}.
84 Easy
84 of 528
Analyzing +4 | -1
If α\alpha and β\beta are the roots of the equation x2+px+2=0{x^2} + px + 2 = 0 and 1α\frac{1}{\alpha } and 1β\frac{1}{\beta } are the roots of the equation 2x2+2qx+1=02{x^2} + 2qx + 1 = 0, then (α−1α)(β−1β)(α+1β)(β+1α)\left( {\alpha - \frac{1}{\alpha }} \right)\left( {\beta - \frac{1}{\beta }} \right)\left( {\alpha + \frac{1}{\beta }} \right)\left( {\beta + \frac{1}{\alpha }} \right) is equal to
View answer and explanation
Correct option: D
94(9−p2)\frac{9}{4}(9 - {p^2})
Option D 94(9−p2)\frac{9}{4}(9 - {p^2}) is correct because, after simplifying the expression using the relationships between roots and coefficients of the given quadratic equations, the result matches this option.
85 Easy
85 of 528
Analyzing +4 | -1
The equation 2cos⁡2(x2)sin⁡2x=x2+1x2,0≤x≤π22{\cos ^2}\left( {\frac{x}{2}} \right){\sin ^2}x = {x^2} + \frac{1}{{{x^2}}},0 \le x \le \frac{\pi }{2} has
View answer and explanation
Correct option: A
No real solution
Option A (No real solution) is correct because the maximum value of the LHS is 2, while the minimum value of the RHS is 2, but they don't attain these values at the same x.
86 Easy
86 of 528
Application +4 | -1
If α,β\alpha, \beta are the roots of the equation ax2+a x^2+ bx+c=0b x+c=0, then the value of α3+β3\alpha^3+\beta^3 is
View answer and explanation
Correct option: C
3abc−b3a3\frac{3 a b c-b^3}{a^3}
Option C (3abc−b3a3\frac{3 a b c-b^3}{a^3}) is correct because it is derived from the formula for α3+β3\alpha^3 + \beta^3 using Vieta's formulas.
87 Easy
87 of 528
Application +4 | -1
The real values of xx for which 372(13)x(13)x>13^{72}\left(\frac{1}{3}\right)^x\left(\frac{1}{3}\right)^{\sqrt{x}}>1, are
View answer and explanation
Correct option: C
x∈[0,64)x \in[0,64)
Option C (x∈[0,64)x \in[0,64)) is correct because it represents the range of real values of x for which the given inequality holds true, considering the domain of the square root function.
88 Easy
88 of 528
Application +4 | -1
The set of values of pp for which the roots of the equation 3x2+2x+p(p−1)=03 x^2+2 x+p(p-1)=0 are of opposite sign, is
View answer and explanation
Correct option: B
(0,1)(0,1)
The correct answer is B ((0,1)) because the roots are of opposite sign when the product of the roots is less than 0, which leads to p(p−1)<0p(p-1) < 0.
89 Easy
89 of 528
Application +4 | -1
The solution of 6+x−x2>0,6 + x - {x^2} > 0, is
View answer and explanation
Correct option: B
−2<x<3- 2 < x < 3
Option B (−2<x<3- 2 < x < 3) is correct because it represents the interval where the quadratic expression is greater than zero, found by factoring and analyzing the inequality.
90 Easy
90 of 528
Understanding +4 | -1
The expression y=ax2+bx+cy=a x^2+b x+c has always the same sign as of 'a' if :
View answer and explanation
Correct option: B
4ac>b24 a c>b^2
Option B (4ac>b24 a c>b^2) is correct because the quadratic expression has the same sign as 'a' when the discriminant is negative, ensuring no real roots and thus no sign change.
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