Solve 2/x=-x+3

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

Solve for x

$$x=1$$

$$x=2$$

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.

$$2=-xx+x\times 3$$

Multiply x and x to get x^{2}.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 2 from both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-3±\sqrt{9+4\left(-2\right)}}{2\left(-1\right)}$$

Multiply 4 times -2.

$$x=\frac{-3±\sqrt{9-8}}{2\left(-1\right)}$$

Add 9 to -8.

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

Take the square root of 1.

$$x=\frac{-3±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-3±1}{-2}$$

Now solve the equation x=\frac{-3±1}{-2} when ± is plus. Add -3 to 1.

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

Divide -2 by -2.

$$x=1$$

Now solve the equation x=\frac{-3±1}{-2} when ± is minus. Subtract 1 from -3.

$$x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x=2$$

The equation is now solved.

$$x=1 x=2$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$2=-xx+x\times 3$$

Multiply x and x to get x^{2}.

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

Swap sides so that all variable terms are on the left hand side.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 3 by -1.

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

Divide 2 by -1.

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

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=2 x=1$$