Solve a^4-20a^2+36=0quada

$$a ^ { 4 } - 20 a ^ { 2 } + 36 = 0 \quad a$$

Solve for a

$$a=3\sqrt{2}\approx 4.242640687$$

$$a=-3\sqrt{2}\approx -4.242640687$$

$$a=-\sqrt{2}\approx -1.414213562$$

$$a=\sqrt{2}\approx 1.414213562$$

Steps Using the Quadratic Formula

Anything times zero gives zero.

$$a^{4}-20a^{2}+36=0$$

Substitute t for a^{2}.

$$t^{2}-20t+36=0$$

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, -20 for b, and 36 for c in the quadratic formula.

$$t=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 1\times 36}}{2}$$

Do the calculations.

$$t=\frac{20±16}{2}$$

Solve the equation t=\frac{20±16}{2} when ± is plus and when ± is minus.

$$t=18 t=2$$

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

$$a=3\sqrt{2} a=-3\sqrt{2} a=\sqrt{2} a=-\sqrt{2}$$