Solve x-4/x=-5

$$x - \frac { 4 } { x } = - 5$$

Solve for x

$$x=\frac{\sqrt{41}-5}{2}\approx 0.701562119$$

$$x=\frac{-\sqrt{41}-5}{2}\approx -5.701562119$$

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.

$$xx-4=-5x$$

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

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

Add 5x to both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{-5±\sqrt{25+16}}{2}$$

Add 25 to 16.

$$x=\frac{-5±\sqrt{41}}{2}$$

Now solve the equation x=\frac{-5±\sqrt{41}}{2} when ± is plus. Add -5 to \sqrt{41}.

$$x=\frac{\sqrt{41}-5}{2}$$

Now solve the equation x=\frac{-5±\sqrt{41}}{2} when ± is minus. Subtract \sqrt{41} from -5.

$$x=\frac{-\sqrt{41}-5}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{41}-5}{2} x=\frac{-\sqrt{41}-5}{2}$$

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.

$$xx-4=-5x$$

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

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

Add 5x to both sides.

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

Add 4 to both sides. Anything plus zero gives itself.

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

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=4+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=4+\frac{25}{4}$$

Add 4 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{41}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{41}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{41}}{2} x+\frac{5}{2}=-\frac{\sqrt{41}}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{\sqrt{41}-5}{2} x=\frac{-\sqrt{41}-5}{2}$$