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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Easy

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Showing 151-180 of 528 questions

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151 Easy
151 of 528
Application +4 | -1
If x2+2ax+b≥c,∀x∈Rx^{2}+2 a x+b \geq c, \forall x \in R, then
View answer and explanation
Correct option: A
b−c≥a2b-c \geq a^{2}
Option A (b−c≥a2b-c \geq a^{2}) is correct because the condition x2+2ax+b≥cx^{2}+2 a x+b \geq c for all xx implies that the discriminant is non-positive.
152 Easy
152 of 528
Application +4 | -1
If equation x2+5bx+8c=0x^{2}+5 b x+8 c=0 does not have two distinct real roots, then minimum value of 5b+8c5 b+8 c is
View answer and explanation
Correct option: D
-1
Option D (-1) is correct because the minimum value of 5b+8c5b + 8c is -1 when the discriminant of the quadratic equation is less than or equal to zero.
153 Easy
153 of 528
Application +4 | -1
Let f(x)f(x) be a quadratic expression such that f(x)<f(x) < 0∀x∈R0 \forall x \in \boldsymbol{R}. If g(x)=f(x)+f′(x)+f′′(x)g(x)=f(x)+f^{\prime}(x)+f^{\prime \prime}(x) then for x∈Rx \in \boldsymbol{R}.
View answer and explanation
Correct option: A
g(x)<0g(x) < 0
Option A (g(x)<0g(x) < 0) is correct because if f(x)f(x) is a quadratic expression always less than 0, then g(x)=f(x)+f′(x)+f′′(x)g(x) = f(x) + f'(x) + f''(x) is also always less than 0.
154 Easy
154 of 528
Analyzing +4 | -1
Let α,β\alpha, \beta be the roots of the equation x2−px+r=0x^2-p x+r=0 and α2,2β\frac{\alpha}{2}, 2 \beta be the roots of the equation x2−qx+r=0x^2-q x+r=0. The value of rr is
View answer and explanation
Correct option: D
29(2p−q)(2q−p)\frac{2}{9}(2 p-q)(2 q-p)
Option D (29(2p−q)(2q−p)\frac{2}{9}(2 p-q)(2 q-p)) is correct because it is derived by expressing the roots α\alpha and β\beta in terms of pp and qq, and then using the relation r=αβr = \alpha\beta.
155 Easy
155 of 528
Application +4 | -1
The value of k(k>0)k(k>0) for which the equations x2+kx+64=0x^2+k x+64=0 and x2−8x+k=0x^2-8 x+k=0, both will have real roots is
View answer and explanation
Correct option: A
16
Option A (16) is correct because for both quadratic equations to have real roots, their discriminants must be non-negative, leading to k=16k=16 as the only positive solution that satisfies both conditions.
156 Easy
156 of 528
Application +4 | -1
If α\alpha and β\beta are the roots of x2+5x+4=0{x^2} + 5x + 4 = 0, then the equation whose roots are α+23,  β+23\frac{{\alpha + 2}}{3},\;\frac{{\beta + 2}}{3} is
View answer and explanation
Correct option: C
9x2+3x−2=09{x^2} + 3x - 2 = 0
The correct answer is C (9x2+3x−2=09{x^2} + 3x - 2 = 0) because it is the quadratic equation whose roots are α+23\frac{{\alpha + 2}}{3} and β+23\frac{{\beta + 2}}{3}, derived from the original equation's roots.
157 Easy
157 of 528
Application +4 | -1
The number of real roots of the following equation ∣x∣2−7∣x∣+12=0|x|^2-7|x|+12=0 is
View answer and explanation
Correct option: D
4
Option D (4) is correct because solving the equation results in four real roots: x = ±3 and x = ±4.
158 Easy
158 of 528
Application +4 | -1
The number of real roots of the equation ∣x∣2−3∣x∣+2=0{\left| x \right|^2} - 3\left| x \right| + 2 = 0 is
View answer and explanation
Correct option: A
4
Option A (4) is correct because solving the equation for |x| yields two positive values, each of which gives two solutions for x (positive and negative).
159 Easy
159 of 528
Analyzing +4 | -1
If the equation x2+px+q=0x^2+p x+q=0 and x2+qx+p=0x^2+q x+p=0, have a common root, then p+q+1=p+q+1= ?
View answer and explanation
Correct option: A
0
Option A (0) is correct because if the two quadratic equations have a common root, substituting that root into both equations and solving leads to p+q+1=0p+q+1=0.
160 Easy
160 of 528
Application +4 | -1
The equation x2−3∣x∣+2=0{x^2} - 3\left| x \right| + 2 = 0 has
View answer and explanation
Correct option: D
Four real roots
The correct answer is D (Four real roots) because by substituting ∣x∣|x| into the equation and solving, we find four distinct real solutions: -2, -1, 1, and 2.
161 Easy
161 of 528
Application +4 | -1
The roots of ∣x−2∣2+∣x−2∣−6=0|x-2|^2+|x-2|-6=0 are
View answer and explanation
Correct option: A
0,4
Option A (0,4) is correct because by substituting |x-2| with a variable and solving the resulting quadratic equation, we find the possible values of x to be 0 and 4.
162 Easy
162 of 528
Application +4 | -1
For what values of kk will the equation x2−2(1+3k)x+7(3+2k)=0x^2-2(1+3 k) x+7(3+2 k)=0 have equal roots?
View answer and explanation
Correct option: B
2,−10/92,-10 / 9
Option B (2,−10/92, -10/9) is correct because these are the values of k that make the discriminant of the quadratic equation equal to zero, resulting in equal roots.
163 Easy
163 of 528
Understanding +4 | -1
Let ff and gg be two real valued functions and SS ={x∣f(x)=0}=\{x \mid f(x)=0\} and T={x∣g(x)=0}T=\{x \mid g(x)=0\}, then S∩TS \cap T represent the set of roots of
View answer and explanation
Correct option: B
f(x)2+g(x)2=0f(x)^2+g(x)^2=0
Option B (f(x)2+g(x)2=0f(x)^2+g(x)^2=0) is correct because f(x)2+g(x)2=0f(x)^2 + g(x)^2 = 0 if and only if f(x)=0f(x) = 0 and g(x)=0g(x) = 0, which means xx is a root of both f(x)f(x) and g(x)g(x).
164 Easy
164 of 528
Analyzing +4 | -1
If three distinct numbers a,a, bb and cc are in GP and the equations ax2+2bx+c=0a{x^2} + 2bx + c = 0 and dx2+2ex+f=0d{x^2} + 2ex + f = 0 have a common root, then which one of the following statements is correct?
View answer and explanation
Correct option: B
da,eb  and  fc  are in AP\frac{d}{a},\frac{e}{b}\,\,\mathrm{and}\,\,\frac{f}{c}\,\,\mathrm{are }\,\mathrm{in }\,\mathrm{AP}
Option B (da,eb and fc are in AP\frac{d}{a},\frac{e}{b}\,\mathrm{and}\,\frac{f}{c}\,\mathrm{are }\,\mathrm{in }\,\mathrm{AP}) is correct because it is derived from the condition that the given quadratic equations have a common root and the coefficients a, b, and c are in GP.
165 Easy
165 of 528
Application +4 | -1
The value of x=2+2+2+……x=\sqrt{2+\sqrt{2+\sqrt{2+\ldots \ldots}}} is:
View answer and explanation
Correct option: C
2
Option C (2) is the correct answer because the expression simplifies to x = 2+x\sqrt{2+x}, which leads to x=2 as the positive solution.

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166 Easy
166 of 528
Application +4 | -1
For a≠ba \ne b, if the equation x2+ax+b=0{x^2} + ax + b = 0 and x2+bx+a=0{x^2} + bx + a = 0 have a common root, then the value of a+ba + b equals
View answer and explanation
Correct option: A
−1- 1
Option A (-1) is correct because substituting the common root into both equations and solving the resulting system leads to a + b = -1.
167 Easy
167 of 528
Application +4 | -1
If a,ba,b and cc are distinct positive real numbers in AP, then the roots of the equation ax2+2bx+c=0a{x^2} + 2bx + c = 0 are
View answer and explanation
Correct option: C
Rational and distinct
The roots are rational and distinct because the discriminant of the quadratic equation is a perfect square, which is greater than zero, indicating distinct rational roots.
168 Easy
168 of 528
Analyzing +4 | -1
If α\alpha and β\beta are the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0 and, if px2+qx+r=0p{x^2} + qx + r = 0 has roots 1−αα\frac{{1 - \alpha }}{\alpha } and 1−ββ\frac{{1 - \beta }}{\beta } , then rr is equal to
View answer and explanation
Correct option: B
a+b+ca + b + c
Option B (a+b+ca + b + c) is correct because substituting the roots into the transformed quadratic equation leads to r = a + b + c.
169 Easy
169 of 528
Analyzing +4 | -1
The equation 34(log⁡2x)2+log⁡2x−54=log⁡x2\frac{3}{4}{\left( {{{\log }_2}x} \right)^2} + {\log _2}x - \frac{5}{4} = {\log _x}\sqrt 2 has
View answer and explanation
Correct option: B
Exactly three real solutions
Option B (Exactly three real solutions) is correct because solving the equation leads to three distinct real values for x.
170 Easy
170 of 528
Analyzing +4 | -1
If the roots of the equations px2+2qx+r=0p x^2+2 q x+r=0 and qx2−2prx+q=0q x^2-2 \sqrt{p r} x+q=0 be real, then
View answer and explanation
Correct option: B
q2=prq^2=p r
Option B (q2=prq^2=p r) is correct. The condition for real roots of both quadratic equations leads to the conclusion that q2q^2 must be equal to prpr.
171 Easy
171 of 528
Application +4 | -1
The solution set of the inequation 4x+32x−5<6,\frac{{4x + 3}}{{2x - 5}} < 6, is
View answer and explanation
Correct option: B
(−∞,5/2)∪(33/8,∞)\left( { - \infty ,5/2} \right) \cup \left( {33/8,\infty } \right)
Option B ((−∞,5/2)∪(33/8,∞)\left( { - \infty ,5/2} \right) \cup \left( {33/8,\infty } \right)) is correct because it represents the solution set of the inequality found by rearranging and solving for x.
172 Easy
172 of 528
Application +4 | -1
The set of values of x for which the inequalities x2−3x−10<0,  10x−x2−16>0{x^2} - 3x - 10 < 0,\;10x - {x^2} - 16 > 0 hold simultaneously, is
View answer and explanation
Correct option: D
(2,5)\left( {2,5} \right)
Option D (2,5)\left( {2,5} \right) is correct because it represents the intersection of the solution sets of both inequalities.
173 Easy
173 of 528
Application +4 | -1
If k∈Rk \in \mathbf{R} lies between the roots of ax2+2bx+a x^2+2 b x+ c=0c=0, then
View answer and explanation
Correct option: B
a2k2+2abk+ac<0a^2 k^2+2 a b k+a c < 0
Option B (a2k2+2abk+ac<0a^2 k^2+2 a b k+a c < 0) is correct because if k lies between the roots, then a*f(k) < 0, where f(k) = ak2 + 2bk + c. Multiplying by 'a' gives a2k2 + 2abk + ac < 0.
174 Easy
174 of 528
Analyzing +4 | -1
If a root of the equation ax2+bx+c=0a{x^2} + bx + c = 0 be reciprocal of a root of the equation a′x2+b′x+c′=0a'{x^2} + b'x + c' = 0, then
View answer and explanation
Correct option: A
(cc′−aa′)2=(ba′−cb′)(ab′−bc′){\left( {cc' - aa'} \right)^2} = \left( {ba' - cb'} \right)\left( {ab' - bc'} \right)
Option A ((cc′−aa′)2=(ba′−cb′)(ab′−bc′){\left( {cc' - aa'} \right)^2} = \left( {ba' - cb'} \right)\left( {ab' - bc'} \right)) is correct because it represents the condition when one root of the first quadratic is the reciprocal of a root of the second quadratic.
175 Easy
175 of 528
Application +4 | -1
If the roots of the given equation (cos⁡p−1)x2+(cos⁡p)x+sin⁡p=0(\cos p - 1){x^2} + (\cos p)x + \sin p = 0 are real, then
View answer and explanation
Correct option: C
p∈(0  ,π)p \in \left( {0\;,\pi } \right)
Option C, p∈(0,π)p \in (0,\pi), is correct because it satisfies the condition for real roots of the quadratic equation, ensuring the discriminant is non-negative.
176 Easy
176 of 528
Application +4 | -1
If the roots of the equation x2+2ax+b=0x^2+2 a x+b=0 are real and distinct and differ by at most 2m2 m, then bb lies in the interval
View answer and explanation
Correct option: C
[a2−m2,a2)\left[a^2-m^2, a^2\right)
Option C ([a2-m2, a2)) is correct because the roots are real and distinct, and their difference is at most 2m, leading to the given interval for b.
177 Easy
177 of 528
Application +4 | -1
The least value of ∣a∣|a| for which sin⁡θ\sin \theta and cosec⁡θ\operatorname{cosec} \theta are the roots of the equation x2+ax+b=0x^{2}+a x+b=0 is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the minimum value of |sin(θ) + cosec(θ)| is 2, which is derived from AM-GM inequality.
178 Easy
178 of 528
Application +4 | -1
If x2−4x+log⁡1/2a=0x^{2}-4 x+\log _{1 / 2} a=0 does not have two distinct real roots, then maximum value of aa is
View answer and explanation
Correct option: B
116\frac{1}{16}
Option B (116\frac{{1}}{{16}}) is correct because the discriminant must be less than or equal to zero for the quadratic to not have two distinct real roots, leading to the inequality a≤116a \le \frac{{1}}{{16}}.
179 Easy
179 of 528
Application +4 | -1
If x=1+ix=1+i is a root of x3−ix+1−i=0x^3-i x+1-i=0, then the quadratic equation whose roots are the remaining two roots of x3−ix+1−i=0x^3-i x+1-i=0 is
View answer and explanation
Correct option: B
x2+(1+i)x+i=0x^2+(1+i) x+i=0
Option B (x2+(1+i)x+i=0x^2+(1+i) x+i=0) is correct because it is the quadratic equation formed by the remaining two roots of the cubic equation after dividing by the factor corresponding to the given root.
180 Easy
180 of 528
Application +4 | -1
The maximum value of the quadratic equation 1−x−3x2=0,x∈R1-x-3 x^2=0, x \in \mathbb{R} is
View answer and explanation
Correct option: D
13/1213 / 12
Option D (13/1213 / 12) is correct because it is the maximum value of the quadratic equation calculated using the formula 4AC−B24A\frac{{4AC-B^2}}{{4A}}.
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