Answer
$x_{1}=\frac{1889+441\sqrt{17}}{512}$
$x_{2}=\frac{1889-441\sqrt{17}}{512}$
Work Step by Step
The quadratic formula will be used, but before doing so, $x^{1/4}=u$ therefore:
$4u^2-9u+4=0$
$u=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where a=4, b=-9, and c=4
$u=\frac{-(-9)\pm\sqrt{(-9)^2-4(4)(4)}}{2(4)}$
$u=\frac{9\pm\sqrt{81-64}}{8}$
$u=\frac{9\pm\sqrt{17}}{8}$
There are two solutions:
First:
$x^{1/4}=\frac{9+\sqrt{17}}{8}$
$(x^{1/4})^4=(\frac{9+\sqrt{17}}{8})^4$
$x=(\frac{9^4+4(9^3)(\sqrt{17})+6(9^2)(\sqrt{17})^2+4(9)(\sqrt{17})^3+(\sqrt{17})^4}{8^4})$
$x=\frac{6561+2916(\sqrt{17})+8262+612(\sqrt{17})+289}{4096}$
$x=\frac{15112+3528\sqrt{17}}{4096}$
$x=\frac{1889+441\sqrt{17}}{512}$
Second:
$x^{1/4}=\frac{9-\sqrt{17}}{8}$
$(x^{1/4})^4=(\frac{9-\sqrt{17}}{8})^4$
$x=(\frac{9^4-4(9^3)(\sqrt{17})+6(9^2)(\sqrt{17})^2-4(9)(\sqrt{17})^3+(\sqrt{17})^4}{8^4})$
$x=\frac{6561-2916(\sqrt{17})+8262-612(\sqrt{17})+289}{4096}$
$x=\frac{15112-3528\sqrt{17}}{4096}$
$x=\frac{1889-441\sqrt{17}}{512}$
