Solve 9x^2+4=(3x^2-1)(2x^2+2)

$$9 x ^ { 2 } + 4 = ( 3 x ^ { 2 } - 1 ) ( 2 x ^ { 2 } + 2 )$$

Solve for x (complex solution)

$$x = \frac{\sqrt{6}}{2} \approx 1.224744871$$

$$x = -\frac{\sqrt{6}}{2} \approx -1.224744871$$

$$x=-\frac{\sqrt{6}i}{3}\approx -0-0.816496581i$$

$$x=\frac{\sqrt{6}i}{3}\approx 0.816496581i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x^{2}-1 by 2x^{2}+2 and combine like terms.

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

Subtract 6x^{4} from both sides.

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

Subtract 4x^{2} from both sides.

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

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

$$5x^{2}+4-6x^{4}=-2$$

Add 2 to both sides.

$$5x^{2}+4-6x^{4}+2=0$$

Add 4 and 2 to get 6.

$$5x^{2}+6-6x^{4}=0$$

Substitute t for x^{2}.

$$-6t^{2}+5t+6=0$$

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 -6 for a, 5 for b, and 6 for c in the quadratic formula.

$$t=\frac{-5±\sqrt{5^{2}-4\left(-6\right)\times 6}}{-6\times 2}$$

Do the calculations.

$$t=\frac{-5±13}{-12}$$

Solve the equation t=\frac{-5±13}{-12} when ± is plus and when ± is minus.

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

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

$$x=-\frac{\sqrt{6}i}{3} x=\frac{\sqrt{6}i}{3} x=-\frac{\sqrt{6}}{2} x=\frac{\sqrt{6}}{2}$$

Solve for x

$$x = -\frac{\sqrt{6}}{2} \approx -1.224744871$$

$$x = \frac{\sqrt{6}}{2} \approx 1.224744871$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x^{2}-1 by 2x^{2}+2 and combine like terms.

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

Subtract 6x^{4} from both sides.

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

Subtract 4x^{2} from both sides.

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

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

$$5x^{2}+4-6x^{4}=-2$$

Add 2 to both sides.

$$5x^{2}+4-6x^{4}+2=0$$

Add 4 and 2 to get 6.

$$5x^{2}+6-6x^{4}=0$$

Substitute t for x^{2}.

$$-6t^{2}+5t+6=0$$

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 -6 for a, 5 for b, and 6 for c in the quadratic formula.

$$t=\frac{-5±\sqrt{5^{2}-4\left(-6\right)\times 6}}{-6\times 2}$$

Do the calculations.

$$t=\frac{-5±13}{-12}$$

Solve the equation t=\frac{-5±13}{-12} when ± is plus and when ± is minus.

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

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

$$x=\frac{\sqrt{6}}{2} x=-\frac{\sqrt{6}}{2}$$