Solve (x^2-4)^2=0

$$( x ^ { 2 } - 4 ) ^ { 2 } = 0$$

Solve for x

$$x=2$$

$$x=-2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-4\right)^{2}.

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

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

$$x^{4}-8x^{2}+16=0$$

Substitute t for x^{2}.

$$t^{2}-8t+16=0$$

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, -8 for b, and 16 for c in the quadratic formula.

$$t=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$t=\frac{8±0}{2}$$

Solutions are the same.

$$t=4$$

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

$$x=-2 x=2$$