Solve (x^2-8)^2-2(x^2-8)+1=0

$$( x ^ { 2 } - 8 ) ^ { 2 } - 2 ( x ^ { 2 } - 8 ) + 1 = 0$$

Solve for x

$$x=3$$

$$x=-3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-8\right)^{2}.

$$\left(x^{2}\right)^{2}-16x^{2}+64-2\left(x^{2}-8\right)+1=0$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{4}-16x^{2}+64-2\left(x^{2}-8\right)+1=0$$

Use the distributive property to multiply -2 by x^{2}-8.

$$x^{4}-16x^{2}+64-2x^{2}+16+1=0$$

Combine -16x^{2} and -2x^{2} to get -18x^{2}.

$$x^{4}-18x^{2}+64+16+1=0$$

Add 64 and 16 to get 80.

$$x^{4}-18x^{2}+80+1=0$$

Add 80 and 1 to get 81.

$$x^{4}-18x^{2}+81=0$$

Substitute t for x^{2}.

$$t^{2}-18t+81=0$$

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, -18 for b, and 81 for c in the quadratic formula.

$$t=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 1\times 81}}{2}$$

Do the calculations.

$$t=\frac{18±0}{2}$$

Solutions are the same.

$$t=9$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for positive t.

$$x=-3 x=3$$