Solve x^4-26x^2+25=0

$${ x }^{ 4 } -26 { x }^{ 2 } +25=0$$

Solve for x

$$x=-5$$

$$x=5$$

$$x=1$$

$$x=-1$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$t^{2}-26t+25=0$$

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, -26 for b, and 25 for c in the quadratic formula.

$$t=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 1\times 25}}{2}$$

Do the calculations.

$$t=\frac{26±24}{2}$$

Solve the equation t=\frac{26±24}{2} when ± is plus and when ± is minus.

$$t=25 t=1$$

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

$$x=5 x=-5 x=1 x=-1$$