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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Easy

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Showing 181-210 of 528 questions

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181 Easy
181 of 528
Application +4 | -1
Let α,β\alpha, \beta be the roots of the equation x2−2x+2=0x^2-\sqrt{2} x+2=0 then α16+β16=\alpha^{16}+\beta^{16}=
View answer and explanation
Correct option: B
−256-256
Option B (-256) is correct because the roots are complex numbers, and raising them to the 16th power simplifies to a real number.
182 Easy
182 of 528
Application +4 | -1
If tan⁡A\tan \mathrm{A} and tan⁡B\tan \mathrm{B} are roots of the quadratic equation x2−px+q=0x^2-p x+q=0, then the value of sin⁡2( A+B)\sin ^2(\mathrm{~A}+\mathrm{B}) is
View answer and explanation
Correct option: C
p2(1−q)2+p2\frac{p^2}{(1-q)^2+p^2}
Option C (p2(1−q)2+p2\frac{{p^2}}{{(1-q)^2+p^2}}) is correct. By using the tangent addition formula and trigonometric identities, we can express sin⁡2(A+B)\sin^2(A+B) in terms of p and q.
183 Easy
183 of 528
Analyzing +4 | -1
The number of irrational solutions of the equation x2+x2+11+x2−x2+11=4,\sqrt {{x^2} + \sqrt {{x^2} + 11} } + \sqrt {{x^2} - \sqrt {{x^2} + 11} } = 4, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because after simplification and substitution, the equation yields two irrational solutions for x.
184 Easy
184 of 528
Application +4 | -1
The minimum value of aa for which a2x2+(2a−1)x+1a^{2} x^{2}+(2 a-1) x+1 is non-negative for any real xx is
View answer and explanation
Correct option: B
14\frac{1}{4}
Option B (14\frac{{1}}{{4}}) is correct because the expression being non-negative implies that the discriminant is non-positive.
185 Easy
185 of 528
Analyzing +4 | -1
If each pair of the equation x2+ax+b=0,  x2+bx+c=0    and   x2+cx+a=0{x^2} + ax + b = 0,\;{x^2} + bx + c = 0\;\,\,\mathrm{and}\,\;{x^2} + cx + a = 0 has a common root, then product of all common roots is
View answer and explanation
Correct option: A
abc\sqrt {abc}
Option A is correct because the product of all common roots is abc\sqrt{abc}, which is derived from Vieta's formulas applied to the given equations.
186 Easy
186 of 528
Analyzing +4 | -1
Sum of the real roots of the equation x2+5∣x∣+6=0x^2+5|x|+6=0 is equal to
View answer and explanation
Correct option: D
Does not exist
Option D (Does not exist) is correct because the equation x2+5∣x∣+6=0x^2 + 5|x| + 6 = 0 has no real roots, as all terms are non-negative, and their sum cannot be zero.
187 Easy
187 of 528
Application +4 | -1
If one root of the equation 8x2−6x−a−3=08{x^2} - 6x - a - 3 = 0 is the square of the other, then the values of aa are
View answer and explanation
Correct option: D
−4,  24- 4,\;24
Option D (−4,  24- 4,\;24) is correct because these are the values of 'a' obtained by setting one root as the square of the other and solving the resulting equations.
188 Easy
188 of 528
Application +4 | -1
α=sin⁡36∘\alpha = \sin 36^\circ is a root of which of the following equation?
View answer and explanation
Correct option: C
16x4−20x2+5=016{x^4} - 20{x^2} + 5 = 0
Option C (16x4−20x2+5=016{x^4} - 20{x^2} + 5 = 0) is correct because substituting α=sin⁡36∘\alpha = \sin 36^\circ into the equation yields a true statement.
189 Easy
189 of 528
Analyzing +4 | -1
x2−11x+ax^2-11 x+a and x2−14x+2ax^2-14 x+2 a will have a common factor, if a=a= ?
View answer and explanation
Correct option: B
0, 24
Option B (0, 24) is correct because setting the two quadratic expressions equal to each other to find a common factor leads to the solutions a=0a=0 and a=24a=24.
190 Easy
190 of 528
Application +4 | -1
The number of rational roots of 81(2x−53x+1)4−45(2x−53x+1)2+4=0,x≠1/381\left(\frac{2 x-5}{3 x+1}\right)^4-45\left(\frac{2 x-5}{3 x+1}\right)^2+4=0, x \neq 1 / 3 is
View answer and explanation
Correct option: D
4
Option D (4) is correct because the equation has four rational roots after solving for x.
191 Easy
191 of 528
Application +4 | -1
If x2/3−7x1/3+10=0{x^{2/3}} - 7{x^{1/3}} + 10 = 0 then the set of values of xx, is
View answer and explanation
Correct option: D
{8,  125}\left\{ {8,\;125} \right\}
Option D ({8,  125}\left\{ {8,\;125} \right\}) is correct because these are the solutions obtained by solving the given equation by substitution.
192 Easy
192 of 528
Application +4 | -1
The value of kk for which the equation
3x2+2x(k2+1)+k2−3k+2=03{x^2} + 2x\left( {{k^2} + 1} \right) + {k^2} - 3k + 2 = 0
has roots of opposite signs, lies in the interval
View answer and explanation
Correct option: C
(1,  2)\left( {1,\;2} \right)
The correct answer is B (A circle) because the locus of z satisfying the given condition is a circle.
193 Easy
193 of 528
Analyzing +4 | -1
sin⁡x+cos⁡x=y2−y+a\sin x + \cos x = {y^2} - y + a has no value of xx for any yy, if 'a' belongs to
View answer and explanation
Correct option: D
(3,∞)\left( {\sqrt {3,} \infty } \right)
Option D ((3∞)\left( {\sqrt {3} \infty } \right)) is correct because it ensures that the range of the quadratic in 'y' does not overlap with the range of sin(x) + cos(x).
194 Easy
194 of 528
Application +4 | -1
Suppose α,β\alpha, \beta are roots of 8x2−10x+3=08 x^2-10 x+3=0, then ∑n=0∞(αx+βn)\sum\limits_{{n=0}}^{\infty}\left(\alpha^x+\beta^n\right) is
View answer and explanation
Correct option: C
6
Option C (6) is correct because by finding the roots and using the formula for the sum of an infinite geometric series, the sum converges to 6.
195 Easy
195 of 528
Application +4 | -1
The value of 7+7−7+7−…\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-\ldots}}}} upto ∞\infty is
View answer and explanation
Correct option: C
3
Option C (3) is correct because after setting up the equation and solving for x, the valid solution is 3, considering the nested radical structure.

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196 Easy
196 of 528
Application +4 | -1
If the difference of the roots of x2−px+q=0x^2-p x+q=0 is unity, then
View answer and explanation
Correct option: C
p2+4q2=(1+2q)2p^2+4 q^2=(1+2 q)^2
The correct answer is C (p2+4q2=(1+2q)2p^2+4 q^2=(1+2 q)^2) because it is derived from the relationship between the roots and coefficients of a quadratic equation and the given condition that the difference of the roots is unity.
197 Easy
197 of 528
Analyzing +4 | -1
Let f(x)=x2+ax+bf\left( x \right) = {x^2} + ax + b, where a,  b∈Ra,\;b \in R. If f(x)=0f\left( x \right) = 0 has all its roots imaginary, then the roots of f(x)+f′(x)+f′′(x)=0f\left( x \right) + f'\left( x \right) + f''\left( x \right) = 0 are
View answer and explanation
Correct option: B
Imaginary
Option B (Imaginary) is correct because if f(x)=0f(x) = 0 has imaginary roots, then f(x)+f′(x)+f′′(x)=0f(x) + f'(x) + f''(x) = 0 also has imaginary roots.
198 Easy
198 of 528
Application +4 | -1
If f(x)=(x−a1)2+(x−a2)2+⋯+(x−an)2f(x)=\left(x-a_{1}\right)^{2}+\left(x-a_{2}\right)^{2}+\cdots+\left(x-a_{n}\right)^{2}, find xx where f(x)f(x) is minimum.
View answer and explanation
Correct option: B
a1+a2+⋯+ann\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}
Option B (a1+a2+⋯+ann\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}) is correct because the minimum value of f(x) occurs at the average of the aia_i values.
199 Easy
199 of 528
Application +4 | -1
If 2−3x−2x2≥0,2 - 3x - 2{x^2} \ge 0, then
View answer and explanation
Correct option: B
−2≤x≤12- 2 \le x \le \frac{1}{2}
Option B (−2≤x≤12- 2 \le x \le \frac{1}{2}) is correct because it represents the interval where the quadratic expression 2−3x−2x22 - 3x - 2{x^2} is non-negative.
200 Easy
200 of 528
Analyzing +4 | -1
If the difference between the roots of x2+ax+b=0 and x2+bx+a=0{x^2} + ax + b = 0\,\mathrm{and}\,{x^2} + bx + a = 0 is same and a≠b,a \ne b, then
View answer and explanation
Correct option: A
a+b+4=0a + b + 4 = 0
Option A (a+b+4=0a + b + 4 = 0) is correct because the condition on the difference of roots leads to this relationship between a and b.
201 Easy
201 of 528
Analyzing +4 | -1
If ax2+bx−ca{x^2} + bx - c is divisible by x2+bx+c,{x^2} + bx + c, then 'a' is a root of the equation
View answer and explanation
Correct option: A
cx2−bx−1=0c{x^2} - bx - 1 = 0
Option C (cx2−bx−1=0c{x^2} - bx - 1 = 0) is correct because 'a' satisfies this equation when ax3+bx+ca{x^3} + bx + c is divisible by x2+bx+c{x^2} + bx + c.
202 Easy
202 of 528
Application +4 | -1
If 10x−1+10−x−1=13,{10^{x - 1}} + {10^{ - x - 1}} = \frac{1}{3}, then xx equals to
View answer and explanation
Correct option: A
±log⁡103\pm {\log _{10}}3
Option A (±log⁡103\pm {\log _{10}}3) is correct because solving the given equation leads to 10x=310^x = 3 or 10x=1310^x = \frac{1}{3}, which gives x=±log⁡103x = \pm {\log_{10}}3.
203 Easy
203 of 528
Application +4 | -1
If the roots of the equation (b−c)x2+(c−a)x+(a−b)=0(b-c) x^2+(c-a) x+(a-b)=0 be equal and a+c=kba+c=k b. Then the value of k=\boldsymbol{k}=
View answer and explanation
Correct option: A
2
Option A (2) is correct because, with equal roots and a+c=kb, solving the quadratic equation leads to k=2.
204 Easy
204 of 528
Application +4 | -1
The roots of the equation (a+b)x2−15+(a−b)x2−15=2a,{(a + \sqrt b )^{{x^2} - 15}} + {(a - \sqrt b )^{{x^2} - 15}} = 2a, where a2−b=1a^2-b=1, are
View answer and explanation
Correct option: B
±4,±14\pm 4, \pm \sqrt{14}
Option B (±4,±14\pm 4, \pm \sqrt{14}) is correct because these values satisfy the given equation after simplification and solving for x.
205 Easy
205 of 528
Application +4 | -1
If α,β,γ\alpha, \beta, \gamma are the roots of the equation, x3+ax2+bx+c=0x^3+a x^2+b x+c=0, then (1−α2)(1−β2)(1−γ2)\left(1-\alpha^2\right)\left(1-\beta^2\right)\left(1-\gamma^2\right) is equal to:
View answer and explanation
Correct option: A
(1+b)2−(a+c)2(1+b)^2-(a+c)^2
Option A ((1+b)2−(a+c)2(1+b)^2-(a+c)^2) is correct because by substituting y=1−x2y = 1-x^2 and simplifying, we can find the product of the roots of the transformed equation.
206 Easy
206 of 528
Analyzing +4 | -1
Let f(x)=ax2+bx+c,  a≠0f\left( x \right) = a{x^2} + bx + c,\;a \ne 0 and Δ=b2−4ac\Delta = {b^2} - 4ac. If α+β,  α2+β2\alpha + \beta ,\;{\alpha ^2} + {\beta ^2} and α3+β3{\alpha ^3} + {\beta ^3} are in GP, then
View answer and explanation
Correct option: C
cΔ=0c\Delta = 0
Option C (cΔ=0c\Delta = 0) is correct because it represents the condition when α+β,;α2+β2\alpha + \beta ,\\;{\alpha ^2} + {\beta ^2} and α3+β3{\alpha ^3} + {\beta ^3} are in GP, which implies either c=0 or the discriminant is zero.
207 Easy
207 of 528
Analyzing +4 | -1
If two equations x2+a2=1−2ax{x^2} + {a^2} = 1 - 2ax and x2+b2=1−2bx{x^2} + {b^2} = 1 - 2bx have only one common root, then
View answer and explanation
Correct option: C
a−b=2a - b = 2
Option C (a - b = 2) is correct because by applying the condition for common roots and simplifying the resulting equation, we arrive at a - b = 2.
208 Easy
208 of 528
Analyzing +4 | -1
If the roots of the quadratic equation (4p−p2−5)x2−(2p−1)x+3p=0\left(4 p-p^2-5\right) x^2-(2 p-1) x+3 p=0 lie on either side of unity, the number of integral values of pp is
View answer and explanation
Correct option: B
2
The correct answer is B (2) because the number of integral values of p is found by analyzing the condition where 1 lies between the roots of the quadratic equation.
209 Easy
209 of 528
Application +4 | -1
The α,β\alpha ,\beta are the roots of the equation x2+ax+b=0{x^2} + ax + b = 0, then 1α2+1β2\frac{1}{{{\alpha ^2}}} + \frac{1}{{{\beta ^2}}} is equal to
View answer and explanation
Correct option: A
a2−2bb2\frac{{{a^2} - 2b}}{{{b^2}}}
Option A is correct because it correctly applies the relationship between roots and coefficients of a quadratic equation to find the value of the expression.
210 Easy
210 of 528
Application +4 | -1
If 32+72i\frac{3}{2} + \frac{7}{2}i is a solution of the equation ax2−6x+b=0a{x^2} - 6x + b = 0, where aa and bb are real numbers, then the value of a+ba + b is equal to
View answer and explanation
Correct option: D
31
Option D (31) is correct because finding the conjugate root and using Vieta's formulas leads to a = 2 and b = 29, so a + b = 31.
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