Solve 4/x=x+3

$$\frac { 4 } { x } = x + 3$$

Solve for x

$$x=-4$$

$$x=1$$

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.

$$4=xx+x\times 3$$

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

$$4=x^{2}+x\times 3$$

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

$$x^{2}+x\times 3=4$$

Subtract 4 from both sides.

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

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

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

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

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

Square 3.

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

Multiply -4 times -4.

$$x=\frac{-3±\sqrt{9+16}}{2}$$

Add 9 to 16.

$$x=\frac{-3±\sqrt{25}}{2}$$

Take the square root of 25.

$$x=\frac{-3±5}{2}$$

Now solve the equation x=\frac{-3±5}{2} when ± is plus. Add -3 to 5.

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

Divide 2 by 2.

$$x=1$$

Now solve the equation x=\frac{-3±5}{2} when ± is minus. Subtract 5 from -3.

$$x=-\frac{8}{2}$$

Divide -8 by 2.

$$x=-4$$

The equation is now solved.

$$x=1 x=-4$$

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.

$$4=xx+x\times 3$$

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

$$4=x^{2}+x\times 3$$

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

$$x^{2}+x\times 3=4$$

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.

$$x^{2}+3x=4$$

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.

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

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

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

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

$$x^{2}+3x+\frac{9}{4}=\frac{25}{4}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

$$x=1 x=-4$$