Question

solve this question please ​

Answer 1

Step-by-step explanation:

Given :-

x= √2+1

To find :-

Find the value of the following :

i) x^1+(1/x^2) ii)x^3+(1/x^3)

Solution:-

Given that :

x =√2+1---------(1)

1/x = 1/(√2+1)

The denominator =√2+1

Rationalising factor of√2+1 is √2-1

On Rationalising the denominator then

=> 1/x = [1/(√2+1)]×[(√2-1)/(√2-1)]

=> 1/x = (√2-1)/(√2+1)(√2-1)

=> 1/x = (√2-1)/[(√2)^2-1^2]

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

=> 1/x = (√2-1)/(2-1)

=> 1/x = (√2-1)/1

=> 1/x =√2-1-----------(2)

Now from (1)&(2)

x+1/x = √2+1+√2-1

=> x+(1/x) = (1-1)+(√2+√2)

x+(1/x) = 2√2 -------(3)

On squaring both sides then

=> [x+(1/x)]^2 = (2√2)^2

=> x^2+(1/x)^2+2(x)(1/x) = 8

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

=> x^2+(1/x^2)+2 = 8

=>x^2+(1/x^2) = 8-2

x^2+(1/x^2) = 6---------(4)

and

x^3+(1/x)^3

We know that

a^3+b^3=(a+b)(a^2-ab+b^2)

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

=> x^3+(1/x^3)= (2√2)[x^2+(1/x^2)-1]

=>x^3+(1/x^3)= (2√2)(6-1)

=> x^3+(1/x^3)= (2√2)(5)

x^3+(1/x^3)= 10√2

Answer:-

i) The value of x^2+(1)x^2) = 6

ii) The value of x^3+(1/x^3) = 10√2

Used formulae:-

• (a+b)(a-b)=a^2-b^2

• (a+b)^2 = a^2+2ab+b^2

• a^3+b^3=(a+b)(a^2-ab+b^2)

• The Rationalising factor of √a+b = √a-b