Solve x^4+1/4=5/4x^2

$${ x }^{ 4 } + \frac{ 1 }{ 4 } = \frac{ 5 }{ 4 } { x }^{ 2 }$$

Solve for x

$$x=-1$$

$$x=1$$

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

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

Steps Using the Quadratic Formula

Subtract \frac{5}{4}x^{2} from both sides.

$$x^{4}+\frac{1}{4}-\frac{5}{4}x^{2}=0$$

Substitute t for x^{2}.

$$t^{2}-\frac{5}{4}t+\frac{1}{4}=0$$

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, -\frac{5}{4} for b, and \frac{1}{4} for c in the quadratic formula.

$$t=\frac{-\left(-\frac{5}{4}\right)±\sqrt{\left(-\frac{5}{4}\right)^{2}-4\times 1\times \frac{1}{4}}}{2}$$

Do the calculations.

$$t=\frac{\frac{5}{4}±\frac{3}{4}}{2}$$

Solve the equation t=\frac{\frac{5}{4}±\frac{3}{4}}{2} when ± is plus and when ± is minus.

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

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

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