Solve x^4-4x^2+3=0

$${ x }^{ 4 } -4 { x }^{ 2 } +3 = 0$$

Solve for x

$$x=-\sqrt{3}\approx -1.732050808$$

$$x=\sqrt{3}\approx 1.732050808$$

$$x=1$$

$$x=-1$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$t^{2}-4t+3=0$$

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, -4 for b, and 3 for c in the quadratic formula.

$$t=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$t=\frac{4±2}{2}$$

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

$$t=3 t=1$$

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

$$x=\sqrt{3} x=-\sqrt{3} x=1 x=-1$$