Solve x^4-13(x^2-3)=3

$$x ^ { 4 } - 13 ( x ^ { 2 } - 3 ) = 3$$

Solve for x

$$x=-3$$

$$x=3$$

$$x=2$$

$$x=-2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -13 by x^{2}-3.

$$x^{4}-13x^{2}+39=3$$

Subtract 3 from both sides.

$$x^{4}-13x^{2}+39-3=0$$

Subtract 3 from 39 to get 36.

$$x^{4}-13x^{2}+36=0$$

Substitute t for x^{2}.

$$t^{2}-13t+36=0$$

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, -13 for b, and 36 for c in the quadratic formula.

$$t=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 1\times 36}}{2}$$

Do the calculations.

$$t=\frac{13±5}{2}$$

Solve the equation t=\frac{13±5}{2} when ± is plus and when ± is minus.

$$t=9 t=4$$

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

$$x=3 x=-3 x=2 x=-2$$