Solve 4x^4=13x^2-9

$$4 { x }^{ 4 } =13 { x }^{ 2 } -9$$

Solve for x

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

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

$$x=1$$

$$x=-1$$

Steps Using the Quadratic Formula

Subtract 13x^{2} from both sides.

$$4x^{4}-13x^{2}=-9$$

Add 9 to both sides.

$$4x^{4}-13x^{2}+9=0$$

Substitute t for x^{2}.

$$4t^{2}-13t+9=0$$

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 4 for a, -13 for b, and 9 for c in the quadratic formula.

$$t=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 4\times 9}}{2\times 4}$$

Do the calculations.

$$t=\frac{13±5}{8}$$

Solve the equation t=\frac{13±5}{8} when ± is plus and when ± is minus.

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

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

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