Solve x^8+10x^4=9

$$x ^ { 8 } + 10 x ^ { 4 } = 9$$

Solve for x (complex solution)

$$x\in \sqrt{2}\left(\frac{1}{2}+\frac{1}{2}i\right)\sqrt[4]{\sqrt{34}+5},\sqrt{2}\left(\frac{1}{2}-\frac{1}{2}i\right)\sqrt[4]{\sqrt{34}+5},\sqrt{2}\left(-\frac{1}{2}+\frac{1}{2}i\right)\sqrt[4]{\sqrt{34}+5},\sqrt{2}\left(-\frac{1}{2}-\frac{1}{2}i\right)\sqrt[4]{\sqrt{34}+5},-i\sqrt[4]{\sqrt{34}-5},\sqrt[4]{\sqrt{34}-5},i\sqrt[4]{\sqrt{34}-5},-\sqrt[4]{\sqrt{34}-5}$$

Steps Using the Quadratic Formula

Subtract 9 from both sides.

$$x^{8}+10x^{4}-9=0$$

Substitute t for x^{4}.

$$t^{2}+10t-9=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 10 for b, and -9 for c in the quadratic formula.

$$t=\frac{-10±\sqrt{10^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$t=\frac{-10±2\sqrt{34}}{2}$$

Solve the equation t=\frac{-10±2\sqrt{34}}{2} when ± is plus and when ± is minus.

$$t=\sqrt{34}-5 t=-\sqrt{34}-5$$

Since x=t^{4}, the solutions are obtained by solving the equation for each t.

$$x=-i\sqrt[4]{\sqrt{34}-5} x=-\sqrt[4]{\sqrt{34}-5} x=i\sqrt[4]{\sqrt{34}-5} x=\sqrt[4]{\sqrt{34}-5} x=\left(\frac{1}{2}-\frac{1}{2}i\right)\sqrt[4]{4\sqrt{34}+20} x=\left(-\frac{1}{2}-\frac{1}{2}i\right)\sqrt[4]{4\sqrt{34}+20} x=\left(-\frac{1}{2}+\frac{1}{2}i\right)\sqrt[4]{4\sqrt{34}+20} x=\left(\frac{1}{2}+\frac{1}{2}i\right)\sqrt[4]{4\sqrt{34}+20}$$

Solve for x

$$x=-\sqrt[4]{\sqrt{34}-5}\approx -0.954759461$$

$$x=\sqrt[4]{\sqrt{34}-5}\approx 0.954759461$$

Steps Using the Quadratic Formula

Subtract 9 from both sides.

$$x^{8}+10x^{4}-9=0$$

Substitute t for x^{4}.

$$t^{2}+10t-9=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 10 for b, and -9 for c in the quadratic formula.

$$t=\frac{-10±\sqrt{10^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$t=\frac{-10±2\sqrt{34}}{2}$$

Solve the equation t=\frac{-10±2\sqrt{34}}{2} when ± is plus and when ± is minus.

$$t=\sqrt{34}-5 t=-\sqrt{34}-5$$

Since x=t^{4}, the solutions are obtained by evaluating x=±\sqrt[4]{t} for positive t.

$$x=\sqrt[4]{\sqrt{34}-5} x=-\sqrt[4]{\sqrt{34}-5}$$