Solve 2(x^2-1)^2-7(x^2-1)+6=0

$$2 ( x ^ { 2 } - 1 ) ^ { 2 } - 7 ( x ^ { 2 } - 1 ) + 6 = 0$$

Solve for x

$$x = -\frac{\sqrt{10}}{2} \approx -1.58113883$$

$$x = \frac{\sqrt{10}}{2} \approx 1.58113883$$

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

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-1\right)^{2}.

$$2\left(\left(x^{2}\right)^{2}-2x^{2}+1\right)-7\left(x^{2}-1\right)+6=0$$

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

$$2\left(x^{4}-2x^{2}+1\right)-7\left(x^{2}-1\right)+6=0$$

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

$$2x^{4}-4x^{2}+2-7\left(x^{2}-1\right)+6=0$$

Use the distributive property to multiply -7 by x^{2}-1.

$$2x^{4}-4x^{2}+2-7x^{2}+7+6=0$$

Combine -4x^{2} and -7x^{2} to get -11x^{2}.

$$2x^{4}-11x^{2}+2+7+6=0$$

Add 2 and 7 to get 9.

$$2x^{4}-11x^{2}+9+6=0$$

Add 9 and 6 to get 15.

$$2x^{4}-11x^{2}+15=0$$

Substitute t for x^{2}.

$$2t^{2}-11t+15=0$$

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 2 for a, -11 for b, and 15 for c in the quadratic formula.

$$t=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\times 15}}{2\times 2}$$

Do the calculations.

$$t=\frac{11±1}{4}$$

Solve the equation t=\frac{11±1}{4} when ± is plus and when ± is minus.

$$t=3 t=\frac{5}{2}$$

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

$$x=\sqrt{3} x=-\sqrt{3} x=\frac{\sqrt{10}}{2} x=-\frac{\sqrt{10}}{2}$$