# What is x^2+x^-2, if x-(1/x)=4?

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First, we'll apply the rule of negative power for the term x^-2:

x^-2 = 1/x^2

To get the squares x^2 and (1/x^2), we'll raise to square the expression x- (1/x).

If we want to raise to square x- (1/x), we'll apply the formula:

(x-y)^2 = x^2 - 2xy + y^2

But x- (1/x) = 4

[x- (1/x)]^2 = 4^2

x^2 -2*x*(1/x) + 1/x^2 = 16

We'll simplify the ratio 2*x*(1/x) = 2

x^2 - 2 + 1/x^2 = 16

We'll add 2 both sides:

x^2 + 1/x^2 = 16 + 2

**The requested sum is: x^2 + 1/x^2 = 18.**

It is known that x-(1/x)=4 and this has to be used to determine the value of x^2+x^-2

Now, (a-b)^2 = a^2 + b^2 - 2*a*b

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

= x^2 + 1/x^2 - 2

Also, (x - 1/x)^2 = 4^2 = 16

This gives x^2 + 1/x^2 - 2 = 16

x^2 + 1/x^2 = 18

The required value of x^2 + 1/x^2 is 18