Solve x^4-19x^2+48=0

$$x ^ { 4 } - 19 x ^ { 2 } + 48 = 0$$

Solve for x

$$x=-4$$

$$x=4$$

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

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

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$t^{2}-19t+48=0$$

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, -19 for b, and 48 for c in the quadratic formula.

$$t=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 1\times 48}}{2}$$

Do the calculations.

$$t=\frac{19±13}{2}$$

Solve the equation t=\frac{19±13}{2} when ± is plus and when ± is minus.

$$t=16 t=3$$

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

$$x=4 x=-4 x=\sqrt{3} x=-\sqrt{3}$$