Question

if x=root 9-root8 find x square +1upon x square​

Answer 1

Step-by-step explanation:

Given:-

x = √9-√8

To find:-

Find the value of x^2+(1/x^2)?

Solution:-

Given that

x =√9-√8

=>x = √(3×3)-√(2×2×2)

=>x = 3-2√2

1/x = 1/(3-2√2)

The Rationalising factor of 3-2√2 is 3+2√2

On Rationalising the denominator then

=> 1/(3-2√2)×(3+2√2)/(3+2√2)

=>(3+2√2)/(3-2√2)(3+2√2)

We know that

(a+b)(a-b)=a^2-b^2

=>(3+2√2)/(3^2-(2√2)^2)

=>(3+2√2)/(9-8)

=>(3+2√2)/1

=>3+2√2

1/x = 3+2√2

we know that

(a+b)^2 =a^2+2ab+b^2

=>a^2+b^2 = (a+b)^2-2ab

=>x^2+(1/x^2)=(x + 1/x)^2-2(x)(1/x)

=>x^2+(1/x)^2=(x+1/x)^2-2

=>x^2+(1/x)^2 = (3-2√2+3+2√2)^2-2

=>x^2+(1/x)^2 = (3+3)^2-2

=>x^2+(1/x^2)=6^2-2

=>x^2+(1/x)^2=36-2

=>x^2+(1/x^2)=34

Answer:-

The value of x^2+(1/x^2) for the given problem is

34

Used formulae:-

• (a+b)^2 =a^2+2ab+b^2

• (a+b)(a-b)=a^2-b^2