Answer
The solutions are $x=\pm1$ and $x=\pm\dfrac{3}{2}$
Work Step by Step
$4x^{4}=13x^{2}-9$
Take all terms to the left side of the equation:
$4x^{4}-13x^{2}+9=0$
Let $u$ be equal to $x^{2}$
If $u=x^{2}$, then $u^{2}=x^{4}$
Rewrite the equation using the new variable $u$:
$4u^{2}-13u+9=0$
Solve by factoring:
$(u-1)(4u-9)=0$
Set both factors equal to $0$ and solve each individual equation for $u$:
$u-1=0$
$u=1$
$4u-9=0$
$4u=9$
$u=\dfrac{9}{4}$
Substitute $u$ back to $u^{2}$ and solve for $x$:
$x^{2}=1$
$\sqrt{x^{2}}=\pm\sqrt{1}$
$x=\pm1$
$x^{2}=\dfrac{9}{4}$
$\sqrt{x^{2}}=\pm\sqrt{\dfrac{9}{4}}$
$x=\pm\dfrac{3}{2}$
The solutions are $x=\pm1$ and $x=\pm\dfrac{3}{2}$
