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JEE • MATHEMATICS • QUADRATIC EQUATIONS

Quadratic equations practice · Easy

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Practice format and chapter quiz

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 421-450 of 528 questions

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421 Easy

421 of 528

Analyzing +4 | -1

The number of elements in the set

\left\{ {x \in R:\left( {\left| x \right| - 3} \right)\left| {x + 4} \right| = 6} \right\}

is equal to:

View answer and explanation

Correct option: A

Option A (2) is correct because solving the equation with absolute values results in two valid solutions for x: -6 and

(\sqrt{73}-1)/2

422 Easy

422 of 528

Application +4 | -1

The solution of the equation

x + \frac{1}{x} = 2

will be

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the solution to the equation is x=1, which is not listed in the options.

423 Easy

423 of 528

Analyzing +4 | -1

Solution of

(x-1)^2(x+4) < 0

View answer and explanation

Correct option: B

(-\infty,-4)

Option B

((-\infty,-4))

is correct because solving the inequality

(x-1)^2(x+4) < 0

results in this interval.

424 Easy

424 of 528

Application +4 | -1

If roots of quadratic equation

\mathrm{ax}^2+\mathrm{bx}+6=0

are equal then minimum value of

(6 a-b+2)

View answer and explanation

Correct option: A

The correct answer is A (1) because the minimum value of the expression is found by completing the square after using the discriminant condition for equal roots.

425 Easy

425 of 528

Application +4 | -1

If the harmonic mean between roots of

(5+\sqrt{2}) x^2-b x+8+2 \sqrt{5}=0

(1) is 4 then

b

equals

View answer and explanation

Correct option: D

4+\sqrt{5}

Option D (

4+\sqrt{5}

) is correct because the harmonic mean formula is applied correctly, and the values are substituted to find the value of b.

426 Easy

426 of 528

Analyzing +4 | -1

Product of the real roots of the equation

{t^2}{x^2} + \left| x \right| + 9 = 0,\left( {t \ne 0} \right)

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the equation has no real roots as the expression

{t^2}{x^2} + \left| x \right| + 9

is always positive for all real values of x.

427 Easy

427 of 528

Analyzing +4 | -1

x

is real, then the value of

\frac{{x + 2}}{{2{x^2} + 3x + 6}}

is equal to

View answer and explanation

Correct option: B

\left( { - \frac{1}{{13}},\;\frac{1}{3}} \right)

Option B

\left( { - \frac{{1}}{{{{13}}}},\\;\frac{{1}}{{{3}}}} \right)

is correct because the range of the expression is found by ensuring the discriminant of the quadratic equation is non-negative.

428 Easy

428 of 528

Analyzing +4 | -1

If the product of

n

positive numbers is

{n^n}

, then their sum is

View answer and explanation

Correct option: D

Never less than

{n^2}

Option D (Never less than

{n^2}

) is correct because, by the AM-GM inequality, the sum of the numbers is always greater than or equal to

{n^2}

429 Easy

429 of 528

Application +4 | -1

If one root of the equation

{x^2} + \left( {\mathrm{l} - 3i} \right)x - 2\left( {\mathrm{l} + i} \right) = 0

- \mathrm{l} + i

, then the other root is

View answer and explanation

Correct option: D

2i

Option D (

2i

) is correct because it is the other root of the quadratic equation, found by using the sum of roots property.

430 Easy

430 of 528

Application +4 | -1

If the roots of the equation

x^2+p^2=8 x+6 p

are real, then

p

belongs to the interval

View answer and explanation

Correct option: C

[-2,8]

Option C ([-2,8]) is correct because it represents the interval of p values for which the discriminant of the given quadratic equation is non-negative, ensuring real roots.

431 Easy

431 of 528

Application +4 | -1

The roots of

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

are

View answer and explanation

Correct option: B

0,4

Option B (

0,4

) is correct because solving the equation with absolute values leads to these two solutions.

432 Easy

432 of 528

Analyzing +4 | -1

How many roots of the equation

x - \frac{1}{{x - 1}} = 1 - \frac{2}{{x - 1}}

have?

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the given equation has no roots, as any potential solution leads to a contradiction.

433 Easy

433 of 528

Analyzing +4 | -1

All the values of

m

for which both roots of the equation

x^2-2 m x+m^2-1=0

are greater than

-2

but less than 4 , lie in the interval

View answer and explanation

Correct option: C

- 1 < m < 3

Option C (

-1 < m < 3

) is correct because it satisfies the condition that both roots of the quadratic equation are greater than -2 and less than 4.

434 Easy

434 of 528

Application +4 | -1

Let

a, b

and

c

be real numbers such that

4 a+2 b+c=0

and

a b>0

. Then the equation

a x^2+b x+c=0

has

View answer and explanation

Correct option: A

real roots

Option A (real roots) is correct because the discriminant D is greater than 0, indicating real roots.

435 Easy

435 of 528

Analyzing +4 | -1

The set of values of

x

which satisfy the inequations

5 x+2<3 x+8

and

\frac{x+2}{x-1} < 4

View answer and explanation

Correct option: D

(-\infty, 1) \cup(2,3)

Option D

((-\infty, 1) \cup(2,3))

is correct because it represents the intersection of the solution sets for both inequalities.

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436 Easy

436 of 528

Application +4 | -1

The value of

k

for which the number 3 lies between the roots of the equation

x^2+(1-2 k) x+\left(k^2-k-2\right)

=0

is given by

View answer and explanation

Correct option: A

2 < k < 5

Option A (2 < k < 5) is correct because it satisfies the condition f(3) < 0, which ensures that 3 lies between the roots of the equation.

437 Easy

437 of 528

Application +4 | -1

a, b, c

are non-zero rational numbers such that

a+b+c=0

, then the roots of the equation

(b+c-a) x^2+(c+a-b) x+(a+b-c)=0

are

View answer and explanation

Correct option: A

both rational

Option A (both rational) is correct because substituting a+b+c=0 into the equation simplifies it, revealing that x=1 is a root, and the other root is also rational since a, b, and c are rational.

438 Easy

438 of 528

Analyzing +4 | -1

\sqrt {3{x^2} - 7x - 30} + \sqrt {2{x^2} - 7x - 5} = x + 5

, then

x

is equal to

View answer and explanation

Correct option: C

Option C (6) is the correct answer because it satisfies the given equation after squaring and simplifying the terms.

439 Easy

439 of 528

Application +4 | -1

\alpha

and

\beta

are the roots of the equation

2 x^2-35 x+2= 0

then the value of

(2 a-35)^3(2 \beta-35)^3

is equal to

View answer and explanation

Correct option: C

Option C (64) is correct because by using the relationship between roots and coefficients, the expression simplifies to a constant value.

440 Easy

440 of 528

Application +4 | -1

The values of '

a

' for which

\left( {{a^2} - 1} \right){x^2} + 2\left( {a - 1} \right)x + 2

is positive for any

x

, are

View answer and explanation

Correct option: D

a < - 3\,\,\mathrm{or}\,\,a > 1

Option D (

a < - 3\,\,\mathrm{or}\,\,a > 1

) is correct because the expression is positive for all

x

a^2 - 1 > 0

and the discriminant is negative, leading to the solution

a < -3

a > 1

441 Easy

441 of 528

Application +4 | -1

Let for

a \neq a_1 \neq 0, f(x)=a x^2+b x+c

g(x)=a_1 x^2+b_1 x+c_1

and

p(x)=f(x)-g(x)

. If

p(x)=0

only for

x=(-1)

and

p(-2)=2

, the value of

p(2)

View answer and explanation

Correct option: A

Option A (18) is correct because by using the given information to determine the quadratic

p(x)

, we can then evaluate

p(2)

to find the answer.

442 Easy

442 of 528

Analyzing +4 | -1

If the roots of

a{x^2} + bx + c = 0\;\mathrm{are}\;\alpha ,\;\beta

and the roots of

A{x^2} + Bx + C = 0\;\mathrm{are}\;\alpha - k,\;\beta - k,\;\mathrm{then}\;\frac{{{B^2} - 4AC}}{{{b^2} - 4ac}}

is equal to

View answer and explanation

Correct option: C

{\left( {\frac{A}{a}} \right)^2}

Option C (

{\left( {\frac{A}{a}} \right)^2}

) is correct because the discriminant ratio is the square of the ratio of the leading coefficients, derived from the relationship between roots and coefficients.

443 Easy

443 of 528

Analyzing +4 | -1

Let

a,\,\,b \in R

be such that the equation

a{x^2} - 2bx + 15 = 0

has a repeated root

\alpha .

\alpha {\text{ }}\,\,{\text{and}}\,\,{\text{ }}\beta

are the roots of the equation

{x^2} - 2bx + 21 = 0,

then

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

is equal to

View answer and explanation

Correct option: B

Option B (58) is correct because by solving the repeated root equation and substituting into the second equation, we find the value of α² + β².

444 Easy

444 of 528

Application +4 | -1

(7+4 \sqrt{3})^{x^2-2}+(7-4 \sqrt{3})^{x^2-2}=14

, then the positive integral value of

x

View answer and explanation

Correct option: A

Option A (1) is correct because solving the equation with exponents leads to x = 1 as the positive integral solution.

445 Easy

445 of 528

Application +4 | -1

x^2+2 a x+10-3 a > 0

for each

x \in \mathbf{R}

, then

View answer and explanation

Correct option: B

-5 < a < 2

Option B (-5 < a < 2) is correct because for the quadratic to be always positive, the discriminant must be negative, leading to the given range for 'a'.

446 Easy

446 of 528

Analyzing +4 | -1

The number of solutions of the equation

\cos x + \left| x \right| = 0

View answer and explanation

Correct option: A

Option A (0) is correct because the graphs of

y = \cos x

and

y = -|x|

do not intersect, indicating no real roots.

447 Easy

447 of 528

Application +4 | -1

The equation whose roots are reciprocal of the roots of the equation

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

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because the coefficients of the quadratic equation are reversed to find the equation with reciprocal roots.

448 Easy

448 of 528

Analyzing +4 | -1

The equation

e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^{x}+1=0

x \in \mathrm{R}

has:

View answer and explanation

Correct option: B

Two solutions and both are negative

Option B (Two solutions and both are negative) is correct because by substituting

e^{x}-e^{-x}=t

and solving the resulting quadratic equation, we find two negative solutions for x.

449 Easy

449 of 528

Application +4 | -1

x = \frac{1}{2}\left( {\sqrt 7 + \frac{1}{{\sqrt 7 }}} \right)

Then,

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

is equal to

View answer and explanation

Correct option: C

Option C (3) is correct because after substituting the value of x and simplifying, the expression evaluates to 3.

450 Easy

450 of 528

Application +4 | -1

\alpha ,\beta

are the roots of

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

then the equation whose roots are

b

and

c

View answer and explanation

Correct option: D

{x^2} + x\left( {\alpha + \beta + \alpha \;\beta } \right) + \alpha \;\beta \left( {\alpha + \beta } \right) = 0

Option D (

{x^2} + x\left( {\alpha + \beta + \alpha \\;\beta } \right) + \alpha \\;\beta \left( {\alpha + \beta } \right) = 0

) is correct because it represents the quadratic equation with roots b and c, derived from the relationships between roots and coefficients of the original equation.

Revision notes and key takeaways

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