Solve 16k^4=40k^2-9

$$16 k ^ { 4 } = 40 k ^ { 2 } - 9$$

Solve for k

$$k = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$k = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$k=\frac{1}{2}=0.5$$

$$k=-\frac{1}{2}=-0.5$$

Steps Using the Quadratic Formula

Subtract 40k^{2} from both sides.

$$16k^{4}-40k^{2}=-9$$

Add 9 to both sides.

$$16k^{4}-40k^{2}+9=0$$

Substitute t for k^{2}.

$$16t^{2}-40t+9=0$$

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 16 for a, -40 for b, and 9 for c in the quadratic formula.

$$t=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 16\times 9}}{2\times 16}$$

Do the calculations.

$$t=\frac{40±32}{32}$$

Solve the equation t=\frac{40±32}{32} when ± is plus and when ± is minus.

$$t=\frac{9}{4} t=\frac{1}{4}$$

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

$$k=\frac{3}{2} k=-\frac{3}{2} k=\frac{1}{2} k=-\frac{1}{2}$$