Solve (x^2-1)(x^2+1)-4(x^2-1)=0

$$( x ^ { 2 } - 1 ) ( x ^ { 2 } + 1 ) - 4 ( x ^ { 2 } - 1 ) = 0$$

Solve for x

$$x=-1$$

$$x=1$$

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

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

Steps Using the Quadratic Formula

Consider \left(x^{2}-1\right)\left(x^{2}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(x^{2}\right)^{2}-1-4\left(x^{2}-1\right)=0$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{4}-1-4\left(x^{2}-1\right)=0$$

Use the distributive property to multiply -4 by x^{2}-1.

$$x^{4}-1-4x^{2}+4=0$$

Add -1 and 4 to get 3.

$$x^{4}+3-4x^{2}=0$$

Substitute t for x^{2}.

$$t^{2}-4t+3=0$$

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, -4 for b, and 3 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{4±2}{2}$$

Solve the equation t=\frac{4±2}{2} when ± is plus and when ± is minus.

$$t=3 t=1$$

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

$$x=\sqrt{3} x=-\sqrt{3} x=1 x=-1$$