Solve (x^2-9)^2=0

$$( x ^ { 2 } - 9 ) ^ { 2 } = 0$$

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}-9\right)^{2}.

$$\left(x^{2}\right)^{2}-18x^{2}+81=0$$

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

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

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.

$$x=-3 x=3$$