Solve (x^2+2)^2-9(x^2+2)+18=0

$$( x ^ { 2 } + 2 ) ^ { 2 } - 9 ( x ^ { 2 } + 2 ) + 18 = 0$$

Solve for x

$$x=-2$$

$$x=2$$

$$x=1$$

$$x=-1$$

Steps Using the Quadratic Formula

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

$$\left(x^{2}\right)^{2}+4x^{2}+4-9\left(x^{2}+2\right)+18=0$$

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

$$x^{4}+4x^{2}+4-9\left(x^{2}+2\right)+18=0$$

Use the distributive property to multiply -9 by x^{2}+2.

$$x^{4}+4x^{2}+4-9x^{2}-18+18=0$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$x^{4}-5x^{2}+4-18+18=0$$

Subtract 18 from 4 to get -14.

$$x^{4}-5x^{2}-14+18=0$$

Add -14 and 18 to get 4.

$$x^{4}-5x^{2}+4=0$$

Substitute t for x^{2}.

$$t^{2}-5t+4=0$$

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, -5 for b, and 4 for c in the quadratic formula.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$t=\frac{5±3}{2}$$

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

$$t=4 t=1$$

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

$$x=2 x=-2 x=1 x=-1$$