Question

If $23=x^4+\frac{1}{x^4}$, then what is the value of $x^2+\frac{1}{x^2}$?

Answer 1

Answer:

$x^2+\frac{1}{x^2}=5$

Step-by-step explanation:

$x^4+\frac{1}{x^4}=23$ is given.

We want to find $x^2+\frac{1}{x^2}$.

If we square the value we want to find, we should wind up with some terms of the left hand side of the given.

$(x^2+\frac{1}{x^2})^2$

Expand:

$x^4+2x^2\frac{1}{x^2}+\frac{1}{x^4}$ (We used the identity: $(x+a)^2=x^2+2xa+a^2)$ for expansion).

Simplify this value:

$x^4+2+\frac{1}{x^4}$

$x^4+\frac{1}{x^4}+2$

We are given that the sum of the first two terms is 23.

This means $(x^2+\frac{1}{x^2})^2=23+2$.

Let's simplify the right hand side.

$(x^2+\frac{1}{x^2})^2=25$

Now to find the value we want we must simply take the square root of both sides.

$x^2+\frac{1}{x^2}=\pm \sqrt{25}$

Simplify the right hand side:

$x^2+\frac{1}{x^2}=\pm 5$

Since $x^2+\frac{1}{x^2}$ is positive for any real value $x$ (that is not zero), then we can conclude $x^2+\frac{1}{x^2}=5$.