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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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Showing 121-150 of 914 questions

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121 Medium
121 of 914
Analyzing +4 | -1
The values of 'a' for which the roots of the equation x2+x+a=0{x^2} + x + a = 0 are real and exceed 'a' are
View answer and explanation
Correct option: C
a<−2a < - 2
Option C (a<−2a < -2) is correct because it satisfies the conditions for the roots to be real and greater than 'a'.
122 Easy
122 of 914
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.
123 Medium
123 of 914
Analyzing +4 | -1
If a∈R−a \in R^{-}and a≠−2a \neq-2, then the equation x2+a∣x∣+1=0x^2+a|x|+1=0
View answer and explanation
Correct option: D
must have either four real roots or no real root
Option D (must have either four real roots or no real root) is correct because the equation can have either four real roots when a < -2 or no real roots depending on the value of the discriminant.
124 Medium
124 of 914
Application +4 | -1
If α,β\alpha, \beta are the roots of x2−6x−2=0.x^2-6 x-2=0. If an=αn−βna_n=\alpha^n-\beta^n for n≥1.n \geq 1. Then the value of a12−2a103a11\frac{a_{12}-2 a_{10}}{3 a_{11}} is
View answer and explanation
Correct option: A
2
Option A (2) is correct because by using the given recurrence relation and the properties of roots, we can simplify the expression to obtain the value 2.
125 Easy
125 of 914
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.
126 Easy
126 of 914
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.
127 Easy
127 of 914
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.
128 Medium
128 of 914
Analyzing +4 | -1
If α,β\alpha, \beta be the positive real roots of the equation x2−px+q=0x^2-p x+q=0, then the value of α1/4+β1/4\alpha^{1 / 4}+\beta^{1 / 4} is (p+6q+4q1/4p+2q)k\left(p+6 \sqrt{q}+4 q^{1 / 4} \sqrt{p+2 \sqrt{q}}\right)^k, where k\mathrm{k} is equal to
View answer and explanation
Correct option: D
14\frac{1}{4}
Option D (1/4) is correct because after simplifying the expression (α1/4+β1/4)4(\alpha^{1/4} + \beta^{1/4})^4, we find that k = 1/4.
129 Easy
129 of 914
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).
130 Easy
130 of 914
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.
131 Easy
131 of 914
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.
132 Easy
132 of 914
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 ].
133 Medium
133 of 914
Analyzing +4 | -1
The largest possible value of the expression
S=sin⁡x1⋅cos⁡x2+sin⁡x2⋅cos⁡x3+…..+sin⁡x99⋅cos⁡x100+sin⁡x100⋅cos⁡x1, where x1,x2,x3,….,x100S=\sin x_1 \cdot \cos x_2+\sin x_2 \cdot \cos x_3+\ldots . .+\sin x_{99} \cdot \cos x_{100}+\sin x_{100} \cdot \cos x_1, \text { where } x_1, x_2, x_3, \ldots ., x_{100} are arbitrary real numbers is
View answer and explanation
Correct option: A
50
Option A (50) is correct because using the inequality 2ab≤a2+b22ab \leq a^2 + b^2, the expression S can be bounded by 50.
134 Easy
134 of 914
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.
135 Medium
135 of 914
Analyzing +4 | -1
The number of solutions of 3x2+6x+7+5x2+10x+14=4−2x−x2,\sqrt {3{x^2} + 6x + 7} + \sqrt {5{x^2} + 10x + 14} = 4 - 2x - {x^2}, is
View answer and explanation
Correct option: A
1
The correct answer is A. By completing the square and analyzing the minimum values of the square roots and the maximum value of the quadratic expression, we find that the only solution is x=-1.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
136 Easy
136 of 914
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.
137 Medium
137 of 914
Analyzing +4 | -1
If roots of the equation ax2+bx+c=0;(a,b,c∈N)a{x^2} + bx + c = 0;\left( {a,b,c \in N} \right) are rational numbers, then which of the following cannot be true?
View answer and explanation
Correct option: B
All a,ba,b and cc are odd
Option B (All a,ba,b and cc are odd) cannot be true because it leads to a contradiction when considering the discriminant and the nature of rational roots.
138 Easy
138 of 914
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.
139 Medium
139 of 914
Analyzing +4 | -1
The set of all the values of 'aa' for which the roots of the equation (a−2)x2−2ax+a=0(a-2) x^2-2 a x+a=0 lies in the interval (−2,1)(-2,1).
View answer and explanation
Correct option: A
a∈[0,89)∪{2}a \in\left[0, \frac{8}{9}\right) \cup\{2\}
The correct answer is A (a∈[0,89)∪{2}a \in\left[0, \frac{8}{9}\right) \cup\{2\}) because it represents the set of all 'a' values for which the roots of the given quadratic equation lie in the interval (-2, 1), considering different cases and conditions.
140 Medium
140 of 914
Analyzing +4 | -1
If the roots of the equation x2−2ax+a2+a−3=0{x^2} - 2ax + {a^2} + a - 3 = 0 are real and less than 3, then
View answer and explanation
Correct option: A
a<2a < 2
Option A (a<2a < 2) is correct because the roots being real and less than 3 imposes this constraint on the value of a.
141 Easy
141 of 914
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.
142 Medium
142 of 914
Application +4 | -1
If x,y,z∈Rx, y, z \in \mathbf{R} then the least value of the expression E=3x2+5y2+4z2−6x+20y−8z−3E=3 x^2+5 y^2+4 z^2-6 x+20 y-8 z-3 is:
View answer and explanation
Correct option: B
-30
Option B (-30) is correct because completing the square for each variable allows us to rewrite the expression as a sum of squares plus a constant, and the minimum value occurs when the squares are zero.
143 Easy
143 of 914
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.
144 Easy
144 of 914
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.
145 Medium
145 of 914
Application +4 | -1
A real root of the equation log⁡4{log⁡2(x+8−x)}=0\log _4\left\{\log _2(\sqrt{x+8}-\sqrt{x})\right\}=0 is:
View answer and explanation
Correct option: A
1
Option A (1) is correct because substituting x=1 into the equation satisfies the condition.
146 Easy
146 of 914
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.
147 Medium
147 of 914
Analyzing +4 | -1
If the equation, x2+bx+45=0(b∈R)x^{2}+b x+45=0(b \in R) has conjugate complex roots and they satisfy ∣z+1∣=210|z+1|=2 \sqrt{10}, then:
]
View answer and explanation
Correct option: A
b2−b=30b^{2}-b=30
Option A (b2−b=30b^{2}-b=30) is correct because by using the given condition ∣z+1∣=210|z+1|=2 \sqrt{10} and the properties of complex conjugate roots, we can find the value of b and then calculate b2−bb^{2}-b.
148 Easy
148 of 914
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.
149 Easy
149 of 914
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.
150 Easy
150 of 914
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.
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