Question

if x minus one upon x is equal to 6 find the value of x square + 1 upon x square​

Answer 1

I hope it will be help full for you

Answer 2

$x^2+\dfrac{1}{x^2}=38$

Step-by-step explanation:

We have,

$x-\dfrac{1}{x} =6$

To find, the value of $x^2+\dfrac{1}{x^2}$ = ?

∴ $x-\dfrac{1}{x} =6$

Squaring both sides, we get

$(x-\dfrac{1}{x})^2 =6^2$

⇒ $(x)^2+(\dfrac{1}{x})^2-2(x)(\dfrac{1}{x})=36$

Using the algebraic identity,

$(a-b)^{2}=a^{2}+b^{2}-2ab$

⇒ $x^2+\dfrac{1}{x^2}-2=36$

⇒ $x^2+\dfrac{1}{x^2}=36+2=38$

⇒ $x^2+\dfrac{1}{x^2}=38$

∴ $x^2+\dfrac{1}{x^2}=38$