Solve 2/x^2+3+1/2/x^2=1

$$\frac { 2 } { x ^ { 2 } + 3 } + \frac { \frac { 1 } { 2 } } { x ^ { 2 } } = 1$$

Solve for x

$$x=1$$

$$x=-1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x^{2}\left(x^{2}+3\right), the least common multiple of x^{2}+3,x^{2}.

$$x^{2}\times 2+\left(x^{2}+3\right)\times \frac{1}{2}=x^{2}\left(x^{2}+3\right)$$

Use the distributive property to multiply x^{2}+3 by \frac{1}{2}.

$$x^{2}\times 2+\frac{1}{2}x^{2}+\frac{3}{2}=x^{2}\left(x^{2}+3\right)$$

Combine x^{2}\times 2 and \frac{1}{2}x^{2} to get \frac{5}{2}x^{2}.

$$\frac{5}{2}x^{2}+\frac{3}{2}=x^{2}\left(x^{2}+3\right)$$

Use the distributive property to multiply x^{2} by x^{2}+3.

$$\frac{5}{2}x^{2}+\frac{3}{2}=x^{4}+3x^{2}$$

Subtract x^{4} from both sides.

$$\frac{5}{2}x^{2}+\frac{3}{2}-x^{4}=3x^{2}$$

Subtract 3x^{2} from both sides.

$$\frac{5}{2}x^{2}+\frac{3}{2}-x^{4}-3x^{2}=0$$

Combine \frac{5}{2}x^{2} and -3x^{2} to get -\frac{1}{2}x^{2}.

$$-\frac{1}{2}x^{2}+\frac{3}{2}-x^{4}=0$$

Substitute t for x^{2}.

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

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, -\frac{1}{2} for b, and \frac{3}{2} for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{\frac{1}{2}±\frac{5}{2}}{-2}$$

Solve the equation t=\frac{\frac{1}{2}±\frac{5}{2}}{-2} when ± is plus and when ± is minus.

$$t=-\frac{3}{2} t=1$$

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

$$x=1 x=-1$$

Solve for x (complex solution)

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

$$x=-\frac{\sqrt{6}i}{2}\approx -0-1.224744871i$$

$$x=-1$$

$$x=1$$