Solve a^4-2a^2=a^2-1

$$a ^ { 4 } - 2 a ^ { 2 } = a ^ { 2 } - 1$$

Solve for a

$$a=\frac{1-\sqrt{5}}{2}\approx -0.618033989$$

$$a=\frac{\sqrt{5}-1}{2}\approx 0.618033989$$

$$a = \frac{\sqrt{5} + 1}{2} \approx 1.618033989$$

$$a=\frac{-\sqrt{5}-1}{2}\approx -1.618033989$$

Steps Using the Quadratic Formula

Subtract a^{2} from both sides.

$$a^{4}-2a^{2}-a^{2}=-1$$

Combine -2a^{2} and -a^{2} to get -3a^{2}.

$$a^{4}-3a^{2}=-1$$

Add 1 to both sides.

$$a^{4}-3a^{2}+1=0$$

Substitute t for a^{2}.

$$t^{2}-3t+1=0$$

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, -3 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{3±\sqrt{5}}{2}$$

Solve the equation t=\frac{3±\sqrt{5}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{5}+3}{2} t=\frac{3-\sqrt{5}}{2}$$

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

$$a=\frac{\sqrt{5}+1}{2} a=-\frac{\sqrt{5}+1}{2} a=-\frac{1-\sqrt{5}}{2} a=\frac{1-\sqrt{5}}{2}$$