Solve 16t^4=32t^2-16

$$16 t ^ { 4 } = 32 t ^ { 2 } - 16$$

Solve for t

$$t=1$$

$$t=-1$$

Steps Using the Quadratic Formula

Subtract 32t^{2} from both sides.

$$16t^{4}-32t^{2}=-16$$

Add 16 to both sides.

$$16t^{4}-32t^{2}+16=0$$

Substitute t for t^{2}.

$$16t^{2}-32t+16=0$$

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 16 for a, -32 for b, and 16 for c in the quadratic formula.

$$t=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 16\times 16}}{2\times 16}$$

Do the calculations.

$$t=\frac{32±0}{32}$$

Solutions are the same.

$$t=1$$

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

$$t=-1 t=1$$