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

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{23}i-1}{4}\approx -0.25-1.198957881i$$

$$x=\frac{-1+\sqrt{23}i}{4}\approx -0.25+1.198957881i$$

Steps Using the Quadratic Formula

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

$$-x^{2}\times 2=x+3$$

Subtract x from both sides.

$$-x^{2}\times 2-x=3$$

Subtract 3 from both sides.

$$-x^{2}\times 2-x-3=0$$

Multiply -1 and 2 to get -2.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Multiply -4 times -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+8\left(-3\right)}}{2\left(-2\right)}$$

Multiply 8 times -3.

$$x=\frac{-\left(-1\right)±\sqrt{1-24}}{2\left(-2\right)}$$

Add 1 to -24.

$$x=\frac{-\left(-1\right)±\sqrt{-23}}{2\left(-2\right)}$$

Take the square root of -23.

$$x=\frac{-\left(-1\right)±\sqrt{23}i}{2\left(-2\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{23}i}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{1±\sqrt{23}i}{-4}$$

Now solve the equation x=\frac{1±\sqrt{23}i}{-4} when ± is plus. Add 1 to i\sqrt{23}.

$$x=\frac{1+\sqrt{23}i}{-4}$$

Divide 1+i\sqrt{23} by -4.

$$x=\frac{-\sqrt{23}i-1}{4}$$

Now solve the equation x=\frac{1±\sqrt{23}i}{-4} when ± is minus. Subtract i\sqrt{23} from 1.

$$x=\frac{-\sqrt{23}i+1}{-4}$$

Divide 1-i\sqrt{23} by -4.

$$x=\frac{-1+\sqrt{23}i}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{23}i-1}{4} x=\frac{-1+\sqrt{23}i}{4}$$

Steps for Completing the Square

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

$$-x^{2}\times 2=x+3$$

Subtract x from both sides.

$$-x^{2}\times 2-x=3$$

Multiply -1 and 2 to get -2.

$$-2x^{2}-x=3$$

Divide both sides by -2.

$$\frac{-2x^{2}-x}{-2}=\frac{3}{-2}$$

Dividing by -2 undoes the multiplication by -2.

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

Divide -1 by -2.

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

Divide 3 by -2.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=-\frac{3}{2}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{3}{2}+\frac{1}{16}$$

Add -\frac{3}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{23}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{23}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{23}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{23}i}{4} x+\frac{1}{4}=-\frac{\sqrt{23}i}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{-1+\sqrt{23}i}{4} x=\frac{-\sqrt{23}i-1}{4}$$