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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Medium

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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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Showing 331-360 of 362 questions

331 Medium

331 of 362

Analyzing +4 | -1

For the equation

{\left| x \right|^2} + \left| x \right| - 6 = 0

View answer and explanation

Correct option: B

There are only two distinct roots

Option B (There are only two distinct roots) is correct because solving the equation results in two possible values for x: 2 and -2.

332 Medium

332 of 362

Understanding +4 | -1

\alpha

and

\beta (\alpha < \beta )

are two distinct roots of the equation

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

then

View answer and explanation

Correct option: C

\alpha < - \frac{b}{{2a}} < \beta

Option C (

\alpha < -\frac{b}{{2a}} < \beta

) is correct because, by Rolle's theorem, there exists a root of the derivative between any two roots of the original equation, and the root of the derivative is -b/2a.

333 Medium

333 of 362

Understanding +4 | -1

For positive real number

a,b,c\;

which one of the following holds?

View answer and explanation

Correct option: A

{a^2} + {b^2} + {c^2} \ge bc + ca + ab

Option B (

\min \left( {p,q} \right) = \frac{1}{2}\left( {P + q - \left| {p - q} \right|} \right)

) is correct because it correctly defines the minimum of two numbers using the absolute value function.

334 Medium

334 of 362

Analyzing +4 | -1

The value of

a

for which the equations

{x^3} + ax + 1 = 0\;\,\mathrm{and\;}{x^4} + a{x^2} + 1 = 0

have a common root, is

View answer and explanation

Correct option: A

- 2

Option A (-2) is correct because substituting x=1 (the common root) into the equation yields a = -2.

335 Medium

335 of 362

Analyzing +4 | -1

For the equation

3 x^2+p x+3=0, p>0

, if one of the roots is the square of the other, then the integral value of

p=

View answer and explanation

Correct option: B

Option B (3) is correct because the condition that one root is the square of the other leads to a relationship between the roots and the coefficients, allowing us to solve for p.

336 Medium

336 of 362

Analyzing +4 | -1

The value of

k

for which the equation

\left( {k - 2} \right){x^2} + 8x + k + 4 = 0

has both roots real, distinct and negative, is

View answer and explanation

Correct option: C

Option C (3) is correct because it satisfies the conditions for the quadratic equation to have real, distinct, and negative roots.

337 Medium

337 of 362

Application +4 | -1

Let

\alpha ,\,\beta

be two roots of the equation

{x^2} + {\left( {20} \right)^{1/4}}x + {\left( 5 \right)^{1/2}} = 0

. Then,

{\alpha ^8} + {\beta ^8}

is equal to

View answer and explanation

Correct option: C

Option C (50) is correct because by using Vieta's formulas and algebraic manipulations, we can find the value of alpha⁸ + beta⁸.

338 Medium

338 of 362

Application +4 | -1

If the equation

2{x^2} + 3x + 5\lambda = 0\;\,\mathrm{and}\,\;{x^2} + 2x + 3\lambda = 0

have a common root, then

\lambda

is equal to

View answer and explanation

Correct option: C

0,\; - 1

The correct answer is C (

0,\; - 1

) because these are the values of

\lambda

for which the given quadratic equations have a common root.

339 Medium

339 of 362

Analyzing +4 | -1

a \in R

and the equation

(a-2)(x-[x])^2+2(x-[x])+a^2=0

(where,

[x]

denotes the greatest integer

\leq x

) has no integral solution and has exactly one solution in

(2,3)

, then

a

lies in the interval

View answer and explanation

Correct option: C

(-1,0)

Option C ((-1,0)) is correct because the given conditions lead to the inequality

a(a+1) < 0

, which is satisfied when

a

lies in the interval

(-1, 0)

340 Medium

340 of 362

Application +4 | -1

\alpha

and

\beta

are the roots of

4 x^2+3 x+7=0

, then the value of

\frac{1}{\alpha^3}+\frac{1}{\beta^3}

View answer and explanation

Correct option: C

225 / 343

Option C (225/343) is correct because it is the result of substituting the sum and product of roots into the simplified expression.

341 Medium

341 of 362

Analyzing +4 | -1

Let

P(x)=0

be the polynomial equation of least possible degree with rational coefficients having

\sqrt[3]{7}+\sqrt[3]{49}

as a root. Then, the product of all the roots of

P(x)=0

View answer and explanation

Correct option: A

Option A (56) is correct because the product of the roots of the derived cubic equation is equal to the constant term, which is 56.

342 Medium

342 of 362

Analyzing +4 | -1

If roots of the equation

x^4-8 x^3+b x^2+c x+16=0

are positive, then

View answer and explanation

Correct option: C

b=24, c=-32

Option D (

b=24, c=-32

) is correct because by using AM ≥ GM, all roots are found to be equal, allowing the determination of b and c by expanding

(x-2)^4

343 Medium

343 of 362

Analyzing +4 | -1

The equations

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

and

{x^3} - 2\;{x^2} + 2\;x - 1 = 0

have 2 roots in common. Then,

a + b

must be equal to

View answer and explanation

Correct option: C

Option C (0) is correct because by finding the roots of the cubic equation and substituting them into the quadratic equation, we can derive that a + b = 0.

344 Medium

344 of 362

Application +4 | -1

If for all real values of

a

one root of the equation

x^2-3 a x+f(a)=0

is double of the other,

f(x)

is equal

View answer and explanation

Correct option: C

2 x^2

Option C

2 x^2

is correct because it represents the function

f(x)

derived from the given condition that one root is double the other for all real values of

a

345 Medium

345 of 362

Application +4 | -1

\alpha

is a root of

x^2+x+1=0

satisfying

(1+\alpha)^7=a+b \alpha+c \alpha^2

, then the ordered triplet

(a, b, c)

View answer and explanation

Correct option: C

(3,3,2)

Option C ((3,3,2)) is correct because it correctly identifies the ordered triplet by using the properties of complex cube roots of unity and substituting into the given equation.

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346 Medium

346 of 362

Application +4 | -1

The roots of the given equation

\left( {p - q} \right){x^2} + \left( {q - r} \right)x + \left( {r - p} \right) = 0

are

View answer and explanation

Correct option: C

\frac{{r - p}}{{p - q}},1

The correct answer is C (

\frac{{r - p}}{{p - q}},1

) because by inspection, x=1 is a root, and the other root can be found using the relationship between roots and coefficients.

347 Medium

347 of 362

Analyzing +4 | -1

The solution set of

\frac{|x+3|+x}{x+2} > 1

View answer and explanation

Correct option: C

(-5,-2) \cup(-1, \infty)

Option C (

(-5,-2) \cup (-1, \infty)

) is correct because it represents the solution set obtained by solving the inequality considering both cases of the absolute value.

348 Medium

348 of 362

Application +4 | -1

\alpha

and

\beta

are roots of the equation,

x^{2}-4 \sqrt{2} k x+2 e^{4 \ln k}-1=0

for some

k

, and

\alpha^{2}+\beta^{2}=66

, then

\alpha^{3}+\beta^{3}

is equal to:

View answer and explanation

Correct option: D

-280 \sqrt{2}

Option D (

-280 \sqrt{2}

) is correct because solving for k and substituting it into the expression for

\alpha^{3}+\beta^{3}

yields

-280 \sqrt{2}

349 Medium

349 of 362

Application +4 | -1

2\alpha = - 1 - i\sqrt 3

and

2\beta = - 1 + i\sqrt 3

, then

5{\alpha ^4} + 5{\beta ^4} + 7{\alpha ^{ - 1}}{\beta ^{ - 1}}

is equal to

View answer and explanation

Correct option: D

Option D (2) is correct because, after substituting the given values of

\alpha

and

\beta

and simplifying, the expression evaluates to 2.

350 Medium

350 of 362

Analyzing +4 | -1

a

and

b

are positive and unequal, then

View answer and explanation

Correct option: A

a^4+b^4 > a^3 b+a b^3

Option A (

a^4+b^4 > a^3 b+a b^3

) is correct because the difference between the two expressions can be factored into a positive term, proving the inequality.

351 Medium

351 of 362

Analyzing +4 | -1

If for

x \in R,\frac{1}{3} < \frac{{{x^2} - 2x + 4}}{{{x^2} + 2x + 4}} < 3,

then

\frac{{{{9.3}^{2x}} - {{6.3}^x} + 4}}{{{{9.3}^{2x}} + {{6.3}^x} + 4}}\;

lies between

View answer and explanation

Correct option: B

\frac{1}{3}

and 3

Option B (

\frac{1}{3}

and 3) is correct because the given expression can be transformed into the form of the inequality, and thus lies between the same bounds.

352 Medium

352 of 362

Application +4 | -1

Let

f(x)=x^2+2 x+5

and

\alpha, \beta

be roots of

f\left(\frac{1}{t}\right)=0

, then

\alpha+\beta=

View answer and explanation

Correct option: A

\frac{-2}{5}

Option A (-2/5) is correct because it is the sum of the roots of the transformed quadratic equation.

353 Medium

353 of 362

Analyzing +4 | -1

p

and

q

are odd prime numbers, then which of the following statements about the roots of the equation

x^2+2 p x-4 q=0

is correct?

View answer and explanation

Correct option: B

Always irrational

Option B (Always irrational) is correct because the discriminant is positive but not a perfect square, ensuring irrational roots.

354 Medium

354 of 362

Analyzing +4 | -1

a, b

and

c

are all distinct, then

a \frac{(x-b)(x-c)}{(a-b)(a-c)}+b \frac{(x-c)(x-a)}{(b-c)(b-a)}+c \frac{(x-a)(x-b)}{(c-a)(c-b)}-x \text { is equal to }

View answer and explanation

Correct option: A

Option A (0) is correct because the given expression is a quadratic polynomial that evaluates to zero at three distinct points, implying it must be identically zero.

355 Medium

355 of 362

Application +4 | -1

The solution set of the inequation

\frac{2}{{\left| {x - 4} \right|}} > 1,\;x \ne 4,

View answer and explanation

Correct option: B

\left( {2,\;4} \right) \cup \left( {4,\;6} \right)

Option B (

\left( {2,\;4} \right) \cup \left( {4,\;6} \right)

) is correct because it represents the interval where the inequality holds true, excluding the point where the expression is undefined.

356 Medium

356 of 362

Analyzing +4 | -1

{\left[ {\;x} \right]^2} = \left[ {x + 2} \right],\;\mathrm{where}\;\left[ x \right] =

the greatest integer less than or equal to

x

,then

x

must be such that

View answer and explanation

Correct option: B

\left[ { - 1,\;0} \right] \cup \left[ {2,\;3} \right]

Option B (

\left[ { - 1,\;0} \right] \cup \left[ {2,\;3} \right]

) is correct because it represents the intervals of x that satisfy the given equation involving the greatest integer function.

357 Medium

357 of 362

Application +4 | -1

If the roots of the equation

\left( {{p^2} + {q^2}} \right){x^2} - 2q\left( {p + r} \right)x + \left( {{q^2} + {r^2}} \right) = 0

be real and equal, then

p,\;q,\;r

will be in

View answer and explanation

Correct option: B

The correct answer is B (GP) because the condition for real and equal roots leads to the relation

q^2 = pr

, which defines a geometric progression.

358 Medium

358 of 362

Analyzing +4 | -1

The minimum value of

f\left( x \right) = {e^{({x^4} - {x^3} + {x^2})}}

View answer and explanation

Correct option: C

Option C (1) is correct because the minimum value of the exponent occurs at x=0, making the minimum value of the function e⁰ = 1.

359 Medium

359 of 362

Analyzing +4 | -1

If for

x > 0, f(x)=\left(2-x^n\right)^{1 / n}

and

g(x)=x^2+r x+s

;

r, s \in R

. It is given that

g(x)-x=0

has imaginary roots, then the number of real roots of the equation

g(g(x))-f(f(x))=0

View answer and explanation

Correct option: D

Zero

Option D (Zero) is correct because g(g(x)) - f(f(x)) is always greater than zero, implying no real roots exist for the equation g(g(x)) - f(f(x)) = 0.

360 Medium

360 of 362

Analyzing +4 | -1

The number of real roots of the equation

\sqrt{x^{2}-4 x+3}+\sqrt{x^{2}-9}=\sqrt{4 x^{2}-14 x+6}

, is

View answer and explanation

Correct option: A

Option A (1) is the correct answer because after simplifying and solving the equation, only one real root satisfies the original equation's domain.

Revision notes and key takeaways

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