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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Mixed difficulty

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271 Easy
271 of 914
Analyzing +4 | -1
The value of ' aa ' for which the equation x3+ax+1=0x^3+a x+1=0 and x4+ax2+1=0x^4+a x^2+1=0, have a common root, is
View answer and explanation
Correct option: B
a=−2a=-2
Option B (a=−2a=-2) is correct because by finding the common root of the two equations and substituting it back, we get a = -2.
272 Easy
272 of 914
Application +4 | -1
If α,β\alpha, \beta are roots x2+px+q=0x^2+p x+q=0, then value of α3+β3\alpha^3+\beta^3 is
View answer and explanation
Correct option: B
3pq−p33 p q-p^3
Option B (3pq−p33pq-p^3) is correct because it is the correct application of the identity for α3+β3\alpha^3 + \beta^3 using Vieta's formulas.
273 Easy
273 of 914
Application +4 | -1
The set of values of aa for which 1 lies between the roots of equation x2−ax−a+3=0x^{2}-a x-a+3=0 is
View answer and explanation
Correct option: D
(2,∞)(2, \infty)
The correct answer is D ((2,∞)(2, \infty)) because for 1 to lie between the roots, f(1) must be less than 0, leading to the inequality a > 2.
274 Easy
274 of 914
Application +4 | -1
The values of xx and yy such that yy satisfy the equation (x,  y∈(x,\;y \in real numbers) x2−xy+y2−4x−4y+16=0{x^2} - xy + {y^2} - 4x - 4y + 16 = 0
View answer and explanation
Correct option: A
4,44,4
Option A (4,4) is correct because the condition for real x leads to y=4, and substituting back gives x=4.
275 Easy
275 of 914
Application +4 | -1
If tan⁡25∘\tan 25^{\circ} and tan⁡20∘\tan 20^{\circ} are roots of the quadratic equation x2+2px+q=0x^2+2 p x+q=0, then 2p−q2 p-q is equal to
View answer and explanation
Correct option: B
-1
Option B (-1) is correct because using the tangent addition formula and Vieta's formulas, we can relate the sum and product of the roots to p and q, and then solve for 2p-q.
276 Easy
276 of 914
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.
277 Easy
277 of 914
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.
278 Easy
278 of 914
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.
279 Easy
279 of 914
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.
280 Easy
280 of 914
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.
281 Medium
281 of 914
Application +4 | -1
If tan⁡α,tan⁡β,tan⁡γ\tan \alpha, \tan \beta, \tan \gamma are the roots of the equation x3−px2−r=0x^3-p x^2-r=0, then the value of (1+tan⁡2α)(1+tan⁡2β)(1+tan⁡2γ)\left(1+\tan ^2 \alpha\right)\left(1+\tan ^2 \beta\right)\left(1+\tan ^2 \gamma\right) is equal to
View answer and explanation
Correct option: B
1+(p−r)21+(p-r)^2
Option B (1+(p−r)21+(p-r)^2) is correct because by using Vieta's formulas to relate the roots to the coefficients of the polynomial, and then expanding the expression (1+tan⁡2α)(1+tan⁡2β)(1+tan⁡2γ)(1+\tan^2 \alpha)(1+\tan^2 \beta)(1+\tan^2 \gamma), we arrive at the result 1+(p−r)21+(p-r)^2.
282 Easy
282 of 914
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.
283 Easy
283 of 914
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.
284 Easy
284 of 914
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).
285 Easy
285 of 914
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.

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286 Easy
286 of 914
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.
287 Easy
287 of 914
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.
288 Easy
288 of 914
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.
289 Easy
289 of 914
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).
290 Medium
290 of 914
Application +4 | -1
If tan⁡A\tan A and tan⁡B\tan B are the roots of the equation x2−px+q=0x^2-p x+q=0, then the value of sin⁡2(A+B)\sin ^2(A+B) is
View answer and explanation
Correct option: A
p2p2+(1−q)2\frac{p^2}{p^2+(1-q)^2}
The correct answer is A (p2p2+(1−q)2\frac{p^2}{p^2+(1-q)^2}) because it is derived by using the tangent addition formula and trigonometric identities.
291 Easy
291 of 914
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.
292 Easy
292 of 914
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.
293 Easy
293 of 914
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.
294 Easy
294 of 914
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.
295 Medium
295 of 914
Analyzing +4 | -1
The function f(x)=x3+ax2+bx+c,a2≤3bf\left( x \right) = {x^3} + a{x^2} + bx + c,{a^2} \le 3b has
View answer and explanation
Correct option: C
No extreme value
Option C (No extreme value) is correct because the condition a2≤3ba^2 \le 3b implies that the derivative has no real roots, hence no extreme values.
296 Easy
296 of 914
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.
297 Easy
297 of 914
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.
298 Easy
298 of 914
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.
299 Easy
299 of 914
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.
300 Easy
300 of 914
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.
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