Solve b^4+23b^2-24=0

$$b ^ { 4 } + 23 b ^ { 2 } - 24 = 0$$

Solve for b

$$b=-1$$

$$b=1$$

Steps Using the Quadratic Formula

Substitute t for b^{2}.

$$t^{2}+23t-24=0$$

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, 23 for b, and -24 for c in the quadratic formula.

$$t=\frac{-23±\sqrt{23^{2}-4\times 1\left(-24\right)}}{2}$$

Do the calculations.

$$t=\frac{-23±25}{2}$$

Solve the equation t=\frac{-23±25}{2} when ± is plus and when ± is minus.

$$t=1 t=-24$$

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

$$b=1 b=-1$$