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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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211 Easy

211 of 528

Application +4 | -1

The quadratic equation in

x

such that the arithmetic mean of its roots is 5 and geometric mean of the roots is 4, is given by

View answer and explanation

Correct option: B

{x^2} - 10x + 16 = 0

Option B (

{x^2} - 10x + 16 = 0

) is correct because it is formed using the given arithmetic and geometric means of the roots.

212 Easy

212 of 528

Application +4 | -1

\sin \alpha ,\cos \alpha

are the roots of the equation

a{x^2} + bx + c = 0,

then

View answer and explanation

Correct option: A

{a^2} - {b^2} + 2ac = 0

Option D (

{a^2} - {b^2} + 2ac = 0

) is correct because by using the relationships between the roots and coefficients of the quadratic equation and the trigonometric identity

\sin^2 \alpha + \cos^2 \alpha = 1

, we can derive the given relation.

213 Easy

213 of 528

Application +4 | -1

If one root of equation

{x^2} + ax + 12 = 0

is 4 while the equation

{x^2} + ax + b = 0

has equal roots, then the value of

b

View answer and explanation

Correct option: B

\frac{{49}}{4}

Option B (

\frac{{49}}{4}

) is correct because the value of b is calculated using the condition for equal roots of a quadratic equation.

214 Easy

214 of 528

Application +4 | -1

Let

\alpha

and

\beta

be the roots of

p x^2+q x+r=0

p \neq 0

. If

p, q, r

are in A.P. and

\frac{1}{\alpha}+\frac{1}{\beta}=4

, then value of

|\alpha-\beta|

View answer and explanation

Correct option: D

\frac{2}{9} \sqrt{13}

Option D (

\frac{2}{9} \sqrt{13}

) is correct because using the given conditions that p, q, r are in A.P. and the relationship between the roots, we can find the value of |

\alpha - \beta

215 Easy

215 of 528

Application +4 | -1

If the roots of the equation

\frac{1}{{x + p}} + \frac{1}{{x + q}} = \frac{1}{r}

are equal in magnitude but opposite in sign, then the product of the roots will be

View answer and explanation

Correct option: B

- \frac{{\left( {{p^2} + {q^2}} \right)}}{2}

Option B (

- \frac{{{\left( {{{{p^2} + {q^2}}}} \right)}}}{{{{2}}}}

) is correct because it is the product of the roots calculated using the given condition that the roots are equal in magnitude but opposite in sign.

216 Easy

216 of 528

Analyzing +4 | -1

\frac{1}{\sqrt{\alpha}}

and

\frac{1}{\sqrt{\beta}}

are the roots of the equation,

a x^{2}+b x+1=0(a \neq 0, a, b, \in R)

, then the equation,

x\left(x+b^{3}\right)+\left(a^{3}-3 a b x\right)=0

as roots :

View answer and explanation

Correct option: A

\alpha^{3 / 2}

and

\beta^{3 / 2}

Option A (

\alpha^{3 / 2}

and

\beta^{3 / 2}

) is correct because by using the relationships between roots and coefficients, we can find the roots of the new equation.

217 Easy

217 of 528

Analyzing +4 | -1

Let

p, q

and

r

be real numbers

(\mathrm{p} \neq \mathrm{q}, \mathrm{r} \neq 0)

, such that the roots of the equation

\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}

are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to.

View answer and explanation

Correct option: B

p^{2}+q^{2}

Option B (

p^{2}+q^{2}

) is correct because by setting the sum of the roots to zero (since they are equal in magnitude but opposite in sign) and using the quadratic formula, we can find the sum of the squares of the roots.

218 Easy

218 of 528

Analyzing +4 | -1

\alpha

and

\beta

are the roots of the quadratic equation,

{x^2} + x\sin \theta - 2\sin \theta = 0,\theta \in \left( {0,\frac{\pi }{2}} \right)

then

\frac{{{\alpha ^{12}} + {\beta ^{12}}}}{{\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}}

is equal to

View answer and explanation

Correct option: A

\frac{{{2^{12}}}}{{{{\left( {\sin \theta + 8} \right)}^{12}}}}

Option A (

\frac{{{2^{12}}}}{{{{\left( {\sin \theta + 8} \right)}^{12}}}}

) is correct because it is the simplified form of the given expression using the relationship between the roots and coefficients of the quadratic equation.

219 Easy

219 of 528

Analyzing +4 | -1

If the equation

{x^2} + px + q = 0

has roots

u

and

υ

where

p,q

are non-zero constants. Then,

View answer and explanation

Correct option: A

q{x^2} + px + 1 = 0

has roots

\frac{1}{u}

and

\frac{1}{v}

Option A (

q{x^2} + px + 1 = 0

has roots

\frac{1}{u}

and

\frac{1}{v}

) is correct because if

u

and

v

are roots of

{x^2} + px + q = 0

, then the equation with roots

\frac{1}{u}

and

\frac{1}{v}

is obtained by substituting

x = \frac{1}{y}

in the original equation.

220 Easy

220 of 528

Understanding +4 | -1

In the equation

{a_4}{x^4} + {a_3}{x^3} + {a_2}{x^2} + {a_1}x + {a_0} = 0

roots of the equation are

{\alpha _i},\;\;i = 1,\;2,\;3,\;4.

Now,

x

is replaced by

x - 1

, now roots of new equation are

View answer and explanation

Correct option: B

{\alpha _i} + 1,\;i = 1,\;2,\;3,\;4

Option B is correct because replacing x with x-1 in the polynomial equation results in roots that are the original roots plus 1.

221 Easy

221 of 528

Analyzing +4 | -1

x^2-3 x+2

be a factor of

x^4-p x^2+q

, then

(p, q)=

View answer and explanation

Correct option: D

(5,4)

Option D ((5,4)) is correct because substituting the roots of

x^2-3x+2=0

into

x^4-px^2+q=0

and solving the resulting system of equations yields

p=5

and

q=4

222 Easy

222 of 528

Application +4 | -1

\alpha + \beta = - 2

and

{\alpha ^3} + {\beta ^3} = - 56

, then the quadratic equation whose roots are

\alpha

and

\beta

View answer and explanation

Correct option: D

{x^2} + 2x - 8 = 0

Option D (

{x^2} + 2x - 8 = 0

) is correct because given

\alpha + \beta = - 2

and

{\alpha ^3} + {\beta ^3} = - 56

, we can find

\alpha\beta = -8

and thus form the quadratic equation.

223 Easy

223 of 528

Application +4 | -1

If domain of

f(x)=\sqrt{x^2+b x+4}

\mathbf{R}

, then maximum possible integral value of

b

View answer and explanation

Correct option: C

Option C (4) is correct because it is the maximum integer value of b for which the discriminant of the quadratic inside the square root is non-positive, ensuring the expression is defined for all real numbers.

224 Easy

224 of 528

Analyzing +4 | -1

The total number of roots of the equation

\left| {x - {x^2} - 1} \right| = \left| {2x - 3 - {x^2}} \right|

View answer and explanation

Correct option: A

Option A (1) is correct because the equation simplifies to a linear equation with only one real root.

225 Easy

225 of 528

Application +4 | -1

The least value of

{5^{\sin x - 1}} + {5^{ - \sin x - 1}}

View answer and explanation

Correct option: C

\frac{2}{5}

Option C

\frac{2}{5}

is correct because it is derived using the AM-GM inequality.

Explore Other Chapters in MATHEMATICS

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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

226 Easy

226 of 528

Application +4 | -1

\tan A

and

\tan B

are the roots of the quadratic equation,

3 x^{2}-10 x-25=0

then the value of

3 \sin ^{2}(A+B)-10 \sin (A+B) \cdot \cos (A+B)-25 \cos ^{2}(A+B)

View answer and explanation

Correct option: B

-25

Option A (

\frac{{3}}{{4}}

) is correct because after solving the given equation for x, substituting x into the expression

\sqrt{{4x^2 - 1}}

yields

\frac{{3}}{{4}}

227 Easy

227 of 528

Analyzing +4 | -1

If the sum of the roots of the quadratic equation

a{x^2} + bx + c = 0

is equal to the sum of the square of their reciprocals, then

\frac{a}{c},\;\frac{b}{a}

and

\frac{c}{b}

are in

View answer and explanation

Correct option: C

Harmonic progression

Option A (Φ) is correct because there is no real solution that satisfies both inequalities simultaneously.

228 Easy

228 of 528

Application +4 | -1

If one root of the equation

x^2-\lambda x+12=0

is even prime while

x^2+\lambda x+\mu=0

has equal roots, then

\mu

View answer and explanation

Correct option: B

Option B (16) is correct because the value of

\mu

is found by first determining

\lambda

from the first equation and then using the condition for equal roots in the second equation.

229 Easy

229 of 528

Application +4 | -1

The roots of the equation

{2^{x + 2}}{3^{3x/\left( {x - 1} \right)}} = 9

are given by

View answer and explanation

Correct option: D

- 2,\;1 - \frac{{\mathrm{log}3}}{{\mathrm{log}2}}

Option D (

- 2,\;1 - \frac{{\mathrm{log}3}}{{\mathrm{log}2}}

) is correct because by taking the logarithm of both sides of the equation and solving for x, we obtain the given roots.

230 Easy

230 of 528

Application +4 | -1

\left\{x \in R:|x-2|=x^2\right\}=

View answer and explanation

Correct option: D

\{1,-2\}

Option D (

\{1,-2\}

) is correct because these are the real solutions obtained by solving the absolute value equation.

231 Easy

231 of 528

Application +4 | -1

The sum of the real roots of the equation

{\left| {x - 2} \right|^2} + \left| {x - 2} \right| - 2 = 0,

View answer and explanation

Correct option: A

Option A (4) is correct because the sum of the real roots obtained from solving the absolute value equation is 4.

232 Easy

232 of 528

Analyzing +4 | -1

Number of integer solutions of

\frac{{x + 2}}{{{x^2} + 1}} > \frac{1}{2}

View answer and explanation

Correct option: D

Option D (3) is correct because solving the inequality yields -1 < x < 3, resulting in three integer solutions: 0, 1, and 2.

233 Easy

233 of 528

Application +4 | -1

The rational values of

a

and

b

a x^2+b x+1=0

\frac{1}{4+\sqrt{3}}

is a root, are

View answer and explanation

Correct option: A

a=13, b=-8

Option A (

a=13, b=-8

) is correct because it is obtained by rationalizing the given root and comparing coefficients of the quadratic equation.

234 Easy

234 of 528

Application +4 | -1

The least value of

n \in \mathbf{N}

for which

(n-4) x^2+8 x

+n+2 > 0 \forall x \in \mathbf{R}

, is

View answer and explanation

Correct option: D

Option D (7) is correct because it is the smallest natural number that satisfies the conditions for the quadratic expression to be always positive.

235 Easy

235 of 528

Application +4 | -1

\alpha \ne \beta

and

{\alpha ^2} = 5\alpha - 3,\;{\beta ^2} = 5\beta - 3

, then the equation having

\frac{\alpha }{\beta }

and

\frac{\beta }{\alpha }

as its roots is

View answer and explanation

Correct option: B

3{x^2} - 19x + 3 = 0

Option B (

3{x^2} - 19x + 3 = 0

) is correct because it represents the quadratic equation with roots

\frac{\alpha}{\beta}

and

\frac{\beta}{\alpha}

derived from the given conditions.

236 Easy

236 of 528

Application +4 | -1

The roots of the quadratic equation

{x^2} - 2\sqrt 3 x - 22 = 0

are

View answer and explanation

Correct option: D

Real, rational and unequal

The roots are Real, rational and unequal because the discriminant is a perfect square (100).

237 Easy

237 of 528

Application +4 | -1

\alpha

and

\beta

the roots of

{x^2} - x - 1 = 0\;\,\mathrm{and}\;{A_n} = {\alpha ^n} + {\beta ^n},\;\,\mathrm{then\;AM}\;\mathrm{of}\;{A_{n - 1}}\,\,\mathrm{and}\;{A_n}

View answer and explanation

Correct option: B

\frac{{{A_{n + 1\;}}}}{2}

Option B (

\frac{{{{A_{n + 1\;}}}}}{{{{2}}}}

) is correct because it is derived by using the properties of roots of the given quadratic equation and simplifying the expression for the arithmetic mean.

238 Easy

238 of 528

Analyzing +4 | -1

If the roots of the equation

q{x^2} + px + q = 0

are complex, where

p,q

are real, then the roots of the equation

{x^2} - 4qx + {p^2} = 0

are

View answer and explanation

Correct option: A

Real and unequal

Option A (Real and unequal) is correct because the discriminant of the second equation is positive, indicating real and unequal roots.

239 Easy

239 of 528

Application +4 | -1

If the product of the roots of the equation

x^2-5 k x+2 e^{4 h x}-1=0

is 31 , then sum of the root is

View answer and explanation

Correct option: D

Option D (10) is correct because solving for k from the product of roots and then finding the sum of the roots gives 10.

240 Easy

240 of 528

Analyzing +4 | -1

x^2+3 x+5=0

and

a x^2+b x+c=0

have

a

common root and

a, b, c \in \mathcal{N}

, the minimum value of

a+b+c

View answer and explanation

Correct option: B

Option B (9) is correct because the minimum value of a+b+c is obtained when the two quadratic equations have both roots in common and a, b, c are natural numbers.

Revision notes and key takeaways

Review QUADRATIC EQUATIONS in your textbook or class notes, then use this set to check your understanding.

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