Question

if x = √7 + √5
then find x2 + 1/x2 ​

Answer 1

Answer:

Step-by-step explanation:

If x = $\sqrt{7}$ + $\sqrt{5}$

Find $x^{2}$ + 1/

Step 1 =

If x = $\sqrt{7}$ + $\sqrt{5}$

Then,

1/x = 1/$\sqrt{7}$ + $\sqrt{5}$

Now rationalize the denominator of number 1/$\sqrt{7}$ + $\sqrt{5}$

Multiply both the numerator and the denominator of the number with the conjugate of the denominator $\sqrt{7}$ + $\sqrt{5}$

The conjugate of the denominator is $\sqrt{7}$ - $\sqrt{5}$

That is,

1/x = 1/$\sqrt{7}$ + $\sqrt{5}$ X

Follow the identity

1/x = $\sqrt{7}$ - $\sqrt{5}$/(

    = $\sqrt{7}$ - $\sqrt{5}$/7 - 5

1/x = $\sqrt{7}$ - $\sqrt{5}$/2

Step 2 = Follow the identity

We know that x = $\sqrt{7}$ + $\sqrt{5}$ and 1/x =  

We have to find $x^{2}$ + 1/

$x^{2}$ = ($\sqrt{7}$ + $\sqrt{5}$)^2

1/$x^{2}$ =  ($\sqrt{7}$ - $\sqrt{5}$/2)^2

So now go according to the question

That is,

= $x^{2}$ + 1/

= ($\sqrt{7}$ + $\sqrt{5}$)^2 + (

= [($\sqrt{7}$)^2 + 2(

= [7 + 2$\sqrt{35}$ + 5] + [7 - 2

= 7 + 5 + 2$\sqrt{35}$ + 7 + 5 - 2

= 12 + 2$\sqrt{35}$/1 + 12 - 2

Take L.C.M of the denominators 1 and 4

That is 4

= 12 X 4 + 2$\sqrt{35}$ X 4/1 + 12 - 2

= 48 + 8$\sqrt{35}$/4 + 12 - 2

= 48 + 12 + $\sqrt{35}$(8 - 2)[Took common] whole divided by 4

= 60 + 6$\sqrt{35}$/4

= 30 + 3$\sqrt{35}$/2($x^{2}$ + 1/$x^{2}$)