Solve -5/x=2x-11

$$- \frac { 5 } { x } = 2 x - 11$$

Solve for x

$$x=5$$

$$x=\frac{1}{2}=0.5$$

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.

$$-5=2xx+x\left(-11\right)$$

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

$$-5=2x^{2}+x\left(-11\right)$$

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

$$2x^{2}+x\left(-11\right)=-5$$

Add 5 to both sides.

$$2x^{2}+x\left(-11\right)+5=0$$

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.

$$2x^{2}-11x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -11 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\times 5}}{2\times 2}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 2\times 5}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-11\right)±\sqrt{121-8\times 5}}{2\times 2}$$

Multiply -8 times 5.

$$x=\frac{-\left(-11\right)±\sqrt{121-40}}{2\times 2}$$

Add 121 to -40.

$$x=\frac{-\left(-11\right)±\sqrt{81}}{2\times 2}$$

Take the square root of 81.

$$x=\frac{-\left(-11\right)±9}{2\times 2}$$

The opposite of -11 is 11.

$$x=\frac{11±9}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{11±9}{4}$$

Now solve the equation x=\frac{11±9}{4} when ± is plus. Add 11 to 9.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{11±9}{4} when ± is minus. Subtract 9 from 11.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=5 x=\frac{1}{2}$$

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.

$$-5=2xx+x\left(-11\right)$$

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

$$-5=2x^{2}+x\left(-11\right)$$

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

$$2x^{2}+x\left(-11\right)=-5$$

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.

$$2x^{2}-11x=-5$$

Divide both sides by 2.

$$\frac{2x^{2}-11x}{2}=-\frac{5}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{11}{2}x=-\frac{5}{2}$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-\frac{5}{2}+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-\frac{5}{2}+\frac{121}{16}$$

Add -\frac{5}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{81}{16}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{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}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{9}{4} x-\frac{11}{4}=-\frac{9}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$x=5 x=\frac{1}{2}$$