# Help with solving $x^2-1= \frac{1}{x^2}$

find all real values of $x$ that satisfy $$x^2 -1= \frac{1}{x^2}$$

I can see that this could resemble a quadratic. I have no idea how to proceed though.

Hint: Substitute $y=x^2$ and then solve for $y$.

• Thank you Peter. Good hint. Got the answer and more importantly understood it! Feb 13, 2015 at 17:40

$$x^2=1+\frac{1}{x^2}$$ $$x^2=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}$$ $$x=\mp \sqrt{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+.}}}}$$ we know that $$1+\frac{1}{1+\frac{1}{1+\frac{1}{1+.}}}=Golden Ratio$$ see http://en.wikipedia.org/wiki/Golden_ratio

So $$x=\mp\sqrt{GoldenRatio}$$ This solution gives you the real roots only

$x \neq 0$ because then the right-hand side of the equation would be undefined.

So we can multiply both sides of the equation by $x^2$.

$$x^2 - 1 = \frac{1}{x^2} \iff x^2(x^2-1) = \frac {x^2}{x^2} \iff x^4 - x^2 -1 = 0$$

Put $u = x^2$ to get $$u^2 - u - 1 = 0$$ Solve for $u$ first, then solve for $x$ given $u$.

turn $x^2 - 1 = \frac1{x^2}$ into $$x^4 - x^2 - 1 = 0$$ this is a quadratic equation in $x^2.$ solve for $x^2$ first.

Apply $x^2$ to both sides to get $x^4 -x^2 = 1$, substitute $u=x^2$ and get $u^2 - u = 1$, and we may complete the square $(u-\frac{1}{2})^2 = \frac{5}{4} \Rightarrow u = \frac{1}{2} \pm \sqrt{\frac{5}{4}}$, and resubstitute $x^2 = \frac{1}{2} \pm\sqrt{\frac{5}{4}}$ and extract square roots to get $x = \pm \sqrt{\frac{1}{2} \pm\sqrt{\frac{5}{4}}}$.

Let $t=x^2$, then $$t-1=\dfrac1t\implies t\neq0\\ t^2-t=1\\ t^2-t-1=0\\ t=\dfrac{1\pm\sqrt{1+4}}{2}\\ x^2=\dfrac{1\pm\sqrt{5}}{2}\\ x=\left\{{ \sqrt{\dfrac{1+\sqrt{5}}{2}},\sqrt{\dfrac{1-\sqrt{5}}{2}},-\sqrt{\dfrac{1+\sqrt{5}}{2}},-\sqrt{\dfrac{1-\sqrt{5}}{2}} }\right\}$$

Follow this chain of reasoning: \begin{align} x^2-1=\frac{1}{x^2} &\Longleftrightarrow t-1=\frac{1}{t}\tag{substitute $t=x^2$}\\[1em] &\Longleftrightarrow t^2-t=1\tag{multiply through by $t$}\\[1em] &\Longleftrightarrow t^2-t-1=0\tag{rearrange}\\[1em] &\Longleftrightarrow t=\frac{1\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)}\tag{quadratic formula}\\[1em] &\Longleftrightarrow t=\frac{1\pm\sqrt{5}}{2}\tag{simplify}\\[1em] &\Longleftrightarrow x^2=\frac{1\pm\sqrt{5}}{2}\tag{substitute $t=x^2$}\\[1em] &\Longleftrightarrow x=\pm\sqrt{\frac{1\pm\sqrt{5}}{2}}.\tag{solve for $x$}\\[1em] \end{align} Now all that is left to do is work through the pluses and minuses to solve for all possible values of $x$. Doing this yields your answers: $$x=\left\{\sqrt{\frac{1+\sqrt{5}}{2}},-\sqrt{\frac{1+\sqrt{5}}{2}},\sqrt{\frac{1-\sqrt{5}}{2}},-\sqrt{\frac{1-\sqrt{5}}{2}}\right\}.$$

Put $x^2 =u,$ we get

$u^2 -u -1 = 0$

u = (Golden ratio related) $\dfrac{1\pm \sqrt 5}{2}$

The four values for bi-quadratic are:

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