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SSC > Quadratic Equations

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

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}

Therefore, Required quadratic equation is

{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}.