Explore popular questions from Quadratic Equations for SSC. This collection covers Quadratic Equations previous year SSC questions hand picked by experienced teachers.

General Intelligence and Reasoning
General Awareness
Quantitative Aptitude
English Comprehension
Q 1.

Correct2

Incorrect-0.5

If {tex} x^{2}-3x+1=0 {/tex} find the value of {tex} x+\frac{1}{x} {/tex}

A

0

3

C

2

D

1

##### Explanation

Given equation is

{tex} x^{2}-3x+1=0 => x^{2}+1=3x {/tex}

= {tex} \frac{x^{2}+1}{x}=3 {/tex}

= {tex} \frac{x^{2}}{x}+\frac{1}{x}=3 {/tex}

Therefore, {tex} x+\frac{1}{x}=3 {/tex}

Q 2.

Correct2

Incorrect-0.5

If {tex} 2x^{2}+12x+18=0 {/tex}, what is the value of x ?

A

3

B

2

-3

D

-2

##### Explanation

Given equation is

{tex} 2x^{2}+12x+18=0 {/tex}

= {tex} x^{2}+6x+9=0 {/tex} [divide by 2]

= {tex} x^{2}+3x+3x+9=0 {/tex} [by factorisation method]

= {tex} x \left(x+3\right)+3 \left(x+3\right)=0 {/tex}

= {tex} \left(x+3\right) \left(x+3\right)=0 {/tex}

= {tex} \left(x+3\right)^{2}=0 {/tex}

= x+3 = 0

= x = -3

Q 3.

Correct2

Incorrect-0.5

If one root of {tex} x^{2}-6kx+5=0 {/tex} is 5, find the value of k.

A

{tex} -\frac{1}{2} {/tex}

B

-1

1

D

2

##### Explanation

Given, one roots of {tex} x^{2}-6kx+5=0 \ is \ 5 {/tex}

Therefore, x = 5 satisfies {tex} x^{2}-6kx+5=0 {/tex}

= {tex} 5^{2}-6 \times k \times 5+5=0 {/tex}

= {tex} 25-30k+5=0 {/tex}

=> 30 - 30k = 0

=> 30k = 30

Therefore, k = 1

Q 4.

Correct2

Incorrect-0.5

If one of the roots of quadratic equation {tex} 7x^{2}-50x+k=0 {/tex} is 7, then what is the value of k?

7

B

1

C

{tex} \frac{50}{7} {/tex}

D

{tex} \frac{7}{50} {/tex}

##### Explanation

Given equation is {tex} 7x^{2}-50x+k=0 {/tex}

Here, a = 7, b = -50, c = k

Since, {tex} \alpha + \beta =\frac{-b}{a} {/tex}

Therefore, {tex} \alpha + \beta =\frac{50}{7} {/tex}

or {tex} \beta =\frac{50}{7}-7 {/tex}

=> {tex} \beta =\frac{1}{7} {/tex} [Because {tex} \alpha = 7 {/tex}(given)]

and {tex} \alpha \beta =\frac{c}{a} {/tex}

or {tex} 7 \times \frac{1}{7}=\frac{k}{7} {/tex}

k = 7

Q 5.

Correct2

Incorrect-0.5

Find the roots of the equation {tex} 2x^{2}-11x+15=0 {/tex}

{tex} 3 \ and \ \frac{5}{2} {/tex}

B

{tex} -3 \ and \ -\frac{5}{2} {/tex}

C

{tex} 5 \ and \ \frac{3}{2} {/tex}

D

{tex} -5 \ and \ -\frac{3}{2} {/tex}

##### Explanation

{tex} 2x^{2} - \left(6x + 5x\right) + 15 = 0 {/tex}

[by factorization method]

= {tex} 2x^{2} - 6x - 5x + 15 = 0 {/tex}

= {tex} 2x \left(x - 3\right) - 5 \left(x - 3\right) = 0 {/tex}

= {tex} \left(2x - 5\right) \left(x - 3\right) = 0 {/tex}

Therefore, {tex} x = \frac{5}{2}, 3 {/tex}

Hence, the roots are {tex} \frac{5}{2}, 3 {/tex}

Q 6.

Correct2

Incorrect-0.5

The quadratic equation whose roots are 3 and -1, is

A

{tex} x^{2}-4x+3=0 {/tex}

{tex} x^{2}-2x-3=0 {/tex}

C

{tex} x^{2}+2x-3=0 {/tex}

D

{tex} x^{2}+4x+3=0 {/tex}

##### Explanation

Given that, the roots of the quadratic equation are 3 and -1.

Let {tex} \alpha = 3 \ and \ \beta = -1 {/tex}

Sum of roots = {tex} \alpha + \beta = 3 - 1 = 2 {/tex}

Product of roots = {tex} \alpha . \beta = \left(3\right) \left(-1\right) = -3 {/tex}

{tex} x^{2} - \left( \alpha + \beta \right) x + \alpha \beta = 0 {/tex}

{tex} x^{2} - \left(2\right)x + \left(-3\right) = 0 {/tex}

{tex} x^{2} - 2x - 3 = 0 {/tex}

Q 7.

Correct2

Incorrect-0.5

{tex} x^{2}+x-20=0; y^{2}-y-30=0 {/tex}

A

If x>y

B

{tex} If \ x \ge y {/tex}

C

lfx

{tex} If \ x \le y {/tex}

##### Explanation

{tex} x^{2}+x-20=0 {/tex} [by factorisation method]

= {tex} x^{2}+5x-4x-20=0 {/tex}

= {tex} x \left(x+5\right)-4 \left(x+5\right)=0 {/tex}

= {tex} \left(x+5\right) \left(x-4\right)=0 {/tex}

Therefore, x = -5 or 4

and {tex} y^{2}-y-30=0 {/tex}

= {tex} y^{2}-6y+5y-30=0 {/tex}

= {tex} y \left(y-6\right)+5 \left(y-6\right)=0 {/tex}

= {tex} \left(y-6\right) \left(y+5\right)=0 {/tex}

Therefore, y = 6 or -5

Hence, {tex} y \ge x \ or \ x \le y {/tex}

Q 8.

Correct2

Incorrect-0.5

{tex} 225x^{2}-4=0; \sqrt{225y}+2=0 {/tex}

A

If x>y

B

{tex} If \ x\ge y {/tex}

C

lfx

If x = y or relation cannot be established

##### Explanation

{tex} 225x^{2}-4=0 {/tex}

=> {tex} 225x^{2}=4 => x^{2}=\frac{4}{225} {/tex}

Therefore, {tex} x=\sqrt{\frac{4}{225}}=\pm \frac{2}{15}, \ i.e., \ \frac{2}{15} \ and -\frac{2}{15} {/tex}

and {tex} \sqrt{225y}+2=0 \ or \sqrt{225y}=-2 {/tex}

On squaring both sides, we get

{tex} \sqrt{\left(225y\right)^{2}} = \left(-2\right)^{2} {/tex}

So the relationship cannot be established because {tex} y = \frac{4}{225} {/tex}

lies between {tex} \frac{2}{15} and -\frac{2}{15} {/tex}

Q 9.

Correct2

Incorrect-0.5

{tex} \frac{4}{\sqrt{x}}+\frac{7}{\sqrt{x}}=\sqrt{x;} {/tex} {tex} y^{2}-\frac{ \left(11\right)^{\frac{5}{2}} }{\sqrt{y}} =0 {/tex}

A

If x>y

B

{tex} If \ x\ge y {/tex}

C

lfx

If x = y or relation cannot be established

##### Explanation

{tex} \frac{4}{\sqrt{x}}+\frac{7}{\sqrt{x}}=\sqrt{x} = \frac{11}{\sqrt{x}} = \sqrt{x}{/tex}

Therefore, x = 11

and {tex} y^{2} - \frac{ \left(11\right)^{5/2} }{\sqrt{y}} = 0 = y^{2} = \frac{ \left(11\right)^{5/2} }{ \left(y\right)^{1/2} } {/tex}

= {tex} y^{2} \times y^{1/2} = \left(11\right)^{5/2} {/tex}

= {tex} \left(y\right)^{5/2} = \left(11\right)^{5/2} {/tex}

Therefore, y = 11

Therefore, x = y

Q 10.

Correct2

Incorrect-0.5

{tex} x^{2}-365=364; y-\sqrt{324}=\sqrt{81} {/tex}

A

If x>y

B

{tex} If \ x \ge y {/tex}

C

lfx

{tex} If \ x \le y {/tex}

##### Explanation

{tex} x^{2} - 365 = 364 {/tex}

=>{tex} x^{2} = 364 + 365 {/tex}

Therefore, {tex} x = \sqrt{729} = \pm 27 {/tex}

and {tex} y - \sqrt{324} = \sqrt{81} => y - 18 = 9 {/tex}

Therefore, y = 27

So, {tex} y\ge x \ or \ x \le y {/tex} because y = 27 and x = -27 and x = 27

Q 11.

Correct2

Incorrect-0.5

{tex} 3x^{2}+8x+4=0; 4y^{2}-19y+12=0 {/tex}

A

If x>y

B

{tex} If \ x\ge y {/tex}

C

lfx

{tex} If \ x \le y {/tex}

##### Explanation

{tex} 3x^{2} + 8x +4 = 0 {/tex}

= {tex} 3x^{2}+6x+2x+4=0 {/tex}

= {tex} 3x \left(x+2\right) + 2 \left(x+2\right) = 0 {/tex}

= {tex} \left(x+2\right) \left(3x+2\right) = 0 {/tex}

Therefore, x = -2, {tex} -\frac{2}{3} {/tex}

and {tex} 4y^{2} - 19y + 12 = 0 {/tex}

=> {tex} 4y^{2} - 16y -3y + 12 = 0 {/tex}

=> {tex} 4y \left(y - 4\right) - 3 \left(y - 4\right) = 0 {/tex}

=> {tex} \left(y - 4\right) \left(4y - 3\right) = 0 {/tex}

Therefore, {tex} y = 4, \frac{3}{4} {/tex}

Hence, y > x or x < y

Q 12.

Correct2

Incorrect-0.5

{tex} x^{2}-x-12=0; y^{2}+5y+6=0 {/tex}

A

If x>y

{tex} If \ x\ge y {/tex}

C

lfx

D

{tex} If \ x \le y {/tex}

##### Explanation

{tex} x^{2} - x - 12 = 0 {/tex}
={tex} x^{2} - 4x + 3x - 12 = 0 {/tex}
={tex} x \left(x-4\right) + 3 \left(x-4\right) = 0 {/tex}
={tex} \left(x-4\right) \left(x+3\right) = 0 {/tex}
Therefore, x = -3, 4
and {tex} y^{2} + 5y + 6 = 0 {/tex}
={tex} y^{2} + 3y + 2y + 6 = 0 {/tex}
={tex} y \left(y+3\right) + 2 \left(y+3\right) = 0 {/tex}
={tex} \left(y + 3\right) \left(y - 3\right) = 0 {/tex}
= y = -3, -2
Therefore, {tex} x \ge y {/tex}
[because x = -3 and y = -3, so x = y and x = 4 and y = -2, hence x > y]

Q 13.

Correct2

Incorrect-0.5

The number or real solution of the equation {tex} x^{2}-3|x|+2=0 {/tex} is:

A

2

4

C

1

D

3

##### Explanation

Given equations is {tex} x^{2}-3|x|+2=0 {/tex}. If x > 0 then lxl = x

=> {tex} x^{2}-3x+2=0 {/tex}

=> {tex} x^{2}-2x-x+2=0 {/tex}

=> {tex} x \left(x-2\right)-1 \left(x-2\right)=0 {/tex}

=> {tex} \left(x-1\right) \left(x-2\right)=0 {/tex}

x = 1,2

If x < 0, then {tex} |x|=-x {/tex}

=> {tex} x^{2}+3x+2=0 {/tex}

=> {tex} x^{2}+2x+x+2=0 {/tex}

=> {tex} x \left(x+2\right)+1 \left(x+2\right)=0 {/tex}

=> {tex} \left(x+1\right) \left(x+2\right)=0 {/tex}

=> x = -1, -2

Hence four solutions are possible.

Q 14.

Correct2

Incorrect-0.5

The coefficient of x in the equation {tex} x^{2}+px+q=0 {/tex} was taken as 17 in place of 13 its roots Were found to be -2 and -15. The roots of the original equation are:

A

3,10

-3 , -10

C

-5, -8

D

None of these

##### Explanation

Let the equation (incorrectly written form)

be {tex} x^{2}+17x+q=0 {/tex}

since, roots are -2, -15

Therefore, q = 30

so, correct equation is {tex} x^{2}+13x+30=0 {/tex}

=> {tex} x^{2}+10x+3x+30=0 {/tex}

=> {tex} \left(x+3\right) \left(x+10\right)=0 {/tex}

=> x = -3, -10

Q 15.

Correct2

Incorrect-0.5

The number which exceeds its positive Square roots by 12 is:

A

9

16

C

25

D

none of these

##### Explanation

Let the required number is x

According to given condition

{tex} x=\sqrt{x}+12 => x-12=\sqrt{x} {/tex}

=> {tex} x^{2}-25x+144=0 {/tex}

=> {tex} x^{2}-16x-9x+144=0 {/tex}

Therefore, x = 16, 9

since x = 9 does not hold the condition

Therefore, x = 16

Q 16.

Correct2

Incorrect-0.5

Let a, b, c be real numbers a {tex} \ne {/tex} 0. If {tex} \alpha {/tex} is a root of {tex} a^{2}x^{2}+bx+c=0, {/tex},{tex} \beta {/tex} is a root of {tex} a^{2}x^{2}-bx-c=0 {/tex} and {tex} 0< \alpha < \beta {/tex} then the equation {tex} a^{2}x^{2}+2bx+2c=0 {/tex} has a root of {tex} \gamma {/tex} that always satisfies:

A

{tex} \gamma = \frac{ \alpha + \beta }{2} {/tex}

B

{tex} \gamma = \alpha + \frac{ \beta }{2} {/tex}

C

{tex} \gamma = \alpha {/tex}

{tex} \alpha < \gamma < \beta {/tex}

##### Explanation

Since, {tex} \alpha {/tex} and {tex} \beta {/tex} are the roots of given equation

Let {tex} f \left(x\right)=a^{2}x^{2}+2bx+2c=0 {/tex}

then {tex} f \left( \alpha \right)=a^{2}a^{2}+2ba+2c=0 {/tex}

={tex} a^{2}a^{2}+2 \left(b \alpha +c\right)=a^{2}a^{2}-2a^{2}a^{2} {/tex}

={tex} -a^{2}a^{2}= -ve {/tex}

and {tex} f \left( \beta \right)=a^{2} \beta ^{2}+2 \left(b \beta +c\right)=a^{2} \beta ^{2}+2a^{2} \beta ^{2} {/tex}

= {tex} 3a^{2} \beta ^{2} = +ve {/tex}

since, {tex} f \left( \alpha \right) {/tex} and {tex} f \left( \beta \right) {/tex} are of opposite signs therefore by theory of equations there lies a root {tex} \gamma {/tex} of the equation {tex} f \left(x\right)=0 {/tex} between {tex} \alpha {/tex} and {tex} \beta {/tex} i.e., a < {tex} \gamma {/tex} < {tex} \beta {/tex}.

Q 17.

Correct2

Incorrect-0.5

The equation {tex} x \left(\frac{3}{4}log_{2}x\right)^{2}+ \left(log_{2}x\right)
-\frac{5}{4}=\sqrt{2} {/tex} has

A

at least one real solution

B

exactly three real solution

exactly one irrational solution.

D

all of the above

##### Explanation

For given equation to be meaningful we must have x > 0 for x > 0, the given equation can be written as {tex} \frac{3}{4} \left(log_{2}x\right)^{2}+log_{2}x-\frac{5}{4} {/tex}

= {tex} logx\sqrt{2}=\frac{1}{2}logx^{2} {/tex}

Putting {tex} t=log_{2}x {/tex} so that {tex} logx^{2}=\frac{1}{t} {/tex}

{tex} \therefore \frac{3}{4}t^{2}+t-\frac{5}{4}=\frac{1}{2} \left(\frac{1}{t}\right) {/tex}

=> {tex} 3t^{3}+4t^{2}-5t-2=0 {/tex}

=> {tex} \left(t-1\right) \left(t+2\right) \left(3t+1\right)=0 {/tex}

{tex} log_{2}x=t=1, -2, -\frac{1}{3} {/tex}

=> {tex} x=2, 2^{-2}, 2^{\frac{1}{3}}\ or\ x=2, \frac{1}{4}, \frac{1}{2^{\frac{1}{3}}} {/tex}

Thus, the given equation has exactly three real solutions out of which exactly one is irrational i.e. {tex} \frac{1}{2^{\frac{1}{3}}} {/tex}

Q 18.

Correct2

Incorrect-0.5

The solution of set of the equation {tex} x log x \left(1-x\right)^{2}=9 {/tex} is

{tex} \{ -2, 4 \} {/tex}

B

{tex} \{ 4 \} {/tex}

C

{tex} \{ 0, -2, 4 \} {/tex}

D

none of these

##### Explanation

We have, {tex} x^{logx \left(1-x\right)^{2} }=9 {/tex}

Taking log on both sides, we get

{tex} x \left(9\right)=log x \left(1-x\right)^{2} \left(\because a^{x}=N => log a N = x \right) {/tex}

=> {tex} 9= \left(1-x\right)^{2} => 1+x^{2}-2x-9=0 {/tex}

=> {tex} x^{2}-2x-8=0 {/tex}

=> {tex} x = -2, 4 {/tex}

=> {tex} x = 4 \left(\because x = 2\right) {/tex}

Q 19.

Correct2

Incorrect-0.5

If x is real the expression {tex} \frac{x+2}{2x^{2}+3x+6} {/tex} takes all values in the interval:

A

{tex} \left(\frac{1}{13}, \frac{1}{3}\right) {/tex}

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

C

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

D

none of these.

##### Explanation

Let given expression be y

i.e., {tex} 4=\frac{x+2}{2x^{2}+3x+6} {/tex}

=> {tex} 2x^{2}y+ \left(3y-1\right)x+ \left(6y-2\right)=0 {/tex}

If y ≠ 0 then {tex} \triangle \ge 0 {/tex} for real x.

i.e., {tex} b^{2}-4ac\ge 0 {/tex}

{tex} \therefore \left(3y-1\right)^{2}-8y \left(6y-2\right)\ge 0 {/tex}

=> {tex} -39y^{2}+10y+1\ge 0 {/tex}

=> {tex} \left(13y+1\right) \left(3y-1\right) \le 0 {/tex}

=> {tex} -\frac{1}{13} \le y \le \frac{1}{3} {/tex}

If y = 0, then x = -2 which is real and this value of y is included in above range.

Q 20.

Correct2

Incorrect-0.5

If x is real, then the maximum and minimum values of the expression {tex} \frac{x^{2}
-3x+4}{x^{2}+3x+4} {/tex} will be:

A

2,1

B

{tex} 5, \frac{1}{5} {/tex}

{tex} 7, \frac{1}{7} {/tex}

D

none of these.

##### Explanation

Let {tex} y=\frac{x^{2}-3x+4}{x^{2}+3x+4} {/tex}

=> {tex} \left(y-1\right)x^{2}+3 \left(y+1\right)x+4 \left(y-1\right)=0 {/tex}

For x is real, {tex} D \ge 0 {/tex}

=> {tex} 9 \left(y+1\right)^{2}-16 \left(y-1\right)^{2}\ge 0 {/tex}

=> {tex} -7y^{2}+50y-7\ge 0 {/tex}

=> {tex} 7y^{2}-50y+7 \le 0 {/tex}

=> {tex} \left(y-7\right) \left(7y-1\right) \le 0 {/tex}

=> {tex} y \le 7\ and\ y \ge \frac{1}{7} => \frac{1}{7} \le y \le 7 {/tex}

Hence maximum value 7 and minimum is {tex} \frac{1}{7} {/tex}.

Q 21.

Correct2

Incorrect-0.5

The number of real solutions of the equation {tex} |x^{2}+4x+3|+2x+5=0 {/tex} are:

A

1

2

C

3

D

4

##### Explanation

We have {tex} |x^{2}+4x+3|+2x+5=0 {/tex}

have two cases arise

Case l: When {tex} x^{2}+4x+3+2x+5=0 {/tex}

=> {tex} x^{2}+6x+8=0 {/tex}

=> {tex} \left(x+2\right) \left(x+4\right)=0 {/tex}

=> {tex} x = -2, -4 {/tex}

x = -2 is not satisfying the condition {tex} x^{2}+4x+3>0 {/tex},

so. x = -4 is the only solution of given equation.

Case II: When {tex} x^{2}+4x+3<0 {/tex}

This gives {tex} - \left(x^{2}+4x+3\right)+2x+5=0 {/tex}

=> {tex} -x^{2}-2x+2=0 {/tex}

=> {tex} x^{2}+2x-2=0 {/tex}

=> {tex} \left(x+1+\sqrt{3}\right) \left(x+1-\sqrt{3}\right)=0 {/tex}

=> {tex} x = -1+\sqrt{3}, -1-\sqrt{3} {/tex}

Hence, {tex} x=- \left(1+\sqrt{3}\right) {/tex} satisfy the given condition.

Since, {tex} x^{2}+4x+3<0 {/tex} while {tex} x= -1+\sqrt{3} {/tex} is not satisfying the condition. Thus number of real solutions are two.

Q 22.

Correct2

Incorrect-0.5

If the expression {tex} \left(mx-1+\frac{1}{x}\right) {/tex} is always nonnegative, then the minimum value of m must be:

A

{tex} -\frac{1}{2} {/tex}

B

0

{tex} \frac{1}{4} {/tex}

D

{tex} \frac{1}{2} {/tex}

##### Explanation

We know that {tex} ax^{2}+bx+c\ge 0 {/tex} if a > 0 and {tex} b^{2}-yac \le 0 {/tex}.

Now {tex} mx-1+\frac{1}{x}=0 => \frac{mx^{2}-x+1}{x}\ge 0 {/tex}

=> {tex} mx^{2}-x+1\ge 0\ and\ x > 0 {/tex}

Now, {tex} mx^{2}-x+1\ge 0,{/tex} if m > 0 and {tex} 1-4 m \le 0{/tex} or if m>o and {tex} m\ge \frac{1}{4} {/tex}

Thus the minimum value of m is {tex} \frac{1}{4} {/tex}.

Q 23.

Correct2

Incorrect-0.5

The value of x in the given equation {tex} 4^{x}-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1} {/tex} is:

A

{tex} \frac{4}{3} {/tex}

{tex} \frac{3}{2} {/tex}

C

{tex} \frac{2}{1} {/tex}

D

{tex} \frac{5}{3} {/tex}

##### Explanation

Given equation is

{tex} 4x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1} {/tex}

= {tex} 2^{2x}+2^{2x-1} = 3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}} {/tex}

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

= {tex} 2^{2x}.\frac{3}{2}=3^{x-\frac{1}{2}}.4 => 2^{2x-3}=3^{x-\frac{3}{2}} {/tex}

taking log:

{tex} \left(2x-3\right)log2 = \left(x-\frac{3}{2}\right)log3 {/tex}

=> {tex} 2x log2 - 3 log2 = xlog3-\frac{3}{2}log3 {/tex}

=> {tex} x log 4 - x log 3 = 3 log^{2} - \frac{3}{2} log 3 {/tex}

= {tex} xlog \left(\frac{4}{3}\right)= log 8 - log 3\sqrt{3} {/tex}

=> {tex} log \left(\frac{4}{3}\right)^{x} = log \frac{8}{3\sqrt{3}} {/tex}

=> {tex} \left(\frac{4}{3}\right)^{x} = \frac{8}{3\sqrt{3}} {/tex}

=> {tex} \left(\frac{4}{3}\right)^{x} = \left(\frac{4}{3}\right)^{\frac{3}{2}} {/tex}

{tex} x = \frac{3}{2} {/tex}

Q 24.

Correct2

Incorrect-0.5

The harmonic mean of the roots of equation {tex} \left(5+\sqrt{2}x^{2}-14+\sqrt{5}\right)x+8+2\sqrt{5}=0 {/tex} is:

A

2

4

C

6

D

8

##### Explanation

Given equation is

{tex} \left(5+\sqrt{2}\right)x^{2}- \left(4+\sqrt{5}\right) x+8 +2\sqrt{5} = 0 {/tex}

Let {tex} x_{1} {/tex} and {tex} x_{2} {/tex} are the root of the equation.

=> {tex} x_{1}+x_{2}=\frac{4+\sqrt{5}}{5+\sqrt{2}} {/tex} ...(i)

and {tex} x_{1}x_{2}=\frac{8+2\sqrt{5}}{5+\sqrt{2}}=\frac{2 \left(4+\sqrt{5}\right) }{5+\sqrt{2}}=2 \left(x_{1}+x_{2}\right) {/tex} ...(ii)

Harmonic mean = {tex} \frac{2x_{1}x_{2}}{x_{1}+x_{2}}=\frac{4 \left(x_{1}+x_{2}\right) }{ \left(x_{1}+x_{2}\right) }=4 {/tex}

Q 25.

Correct2

Incorrect-0.5

For what value of {tex} \lambda {/tex} the sum of the squares of the roots of {tex} x^{2}+ \left(2+\lambda\right)n-\frac{1}{2} \left(1+\lambda\right)=0 {/tex} is minimum?

A

{tex} \frac{3}{2} {/tex}

B

1

{tex} \frac{1}{2} {/tex}

D

{tex} \frac{11}{4} {/tex}

##### Explanation

Given equation is {tex} x^{2}+ \left(2+\lambda\right)x-\frac{1}{2} \left(1+\lambda \right)=0 {/tex}.

Let {tex} \alpha \ and \ \beta {/tex} are the roots of given equations

=> {tex} \alpha + \beta =- \left(2+ \lambda \right) {/tex} and {tex} \alpha \beta = - \left(\frac{1+\lambda}{2}\right) {/tex}

Now, {tex} \alpha ^{2}+ \beta ^{2}= \left( \alpha + \beta \right)^{2}-2 \alpha \beta {/tex}

=> {tex} \alpha ^{2}+ \beta ^{2}= \left[ - \left(2+\lambda\right)^{2}+2\frac{ \left(1+\lambda\right) }{2} \right] {/tex}

=>{tex} \alpha ^{2}+ \beta ^{2}= \lambda^{2}+4+4\lambda+1+\lambda=\lambda^{2}+5\lambda+5 {/tex}

Now we take the option simultaneously

=> It is minimum for {tex} \lambda = \frac{1}{2} {/tex}.