Solve 12/x=11-2x

$$\frac { 12 } { x } = 11 - 2 x$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x=4$$

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.

$$12=x\times 11-2xx$$

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

$$12=x\times 11-2x^{2}$$

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

$$x\times 11-2x^{2}=12$$

Subtract 12 from both sides.

$$x\times 11-2x^{2}-12=0$$

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.

$$-2x^{2}+11x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 11 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-2\right)\left(-12\right)}}{2\left(-2\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-2\right)\left(-12\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-11±\sqrt{121+8\left(-12\right)}}{2\left(-2\right)}$$

Multiply 8 times -12.

$$x=\frac{-11±\sqrt{121-96}}{2\left(-2\right)}$$

Add 121 to -96.

$$x=\frac{-11±\sqrt{25}}{2\left(-2\right)}$$

Take the square root of 25.

$$x=\frac{-11±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-11±5}{-4}$$

Now solve the equation x=\frac{-11±5}{-4} when ± is plus. Add -11 to 5.

$$x=-\frac{6}{-4}$$

Reduce the fraction \frac{-6}{-4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-11±5}{-4} when ± is minus. Subtract 5 from -11.

$$x=-\frac{16}{-4}$$

Divide -16 by -4.

$$x=4$$

The equation is now solved.

$$x=\frac{3}{2} x=4$$

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.

$$12=x\times 11-2xx$$

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

$$12=x\times 11-2x^{2}$$

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

$$x\times 11-2x^{2}=12$$

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.

$$-2x^{2}+11x=12$$

Divide both sides by -2.

$$\frac{-2x^{2}+11x}{-2}=\frac{12}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{11}{-2}x=\frac{12}{-2}$$

Divide 11 by -2.

$$x^{2}-\frac{11}{2}x=\frac{12}{-2}$$

Divide 12 by -2.

$$x^{2}-\frac{11}{2}x=-6$$

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.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-6+\left(-\frac{11}{4}\right)^{2}$$

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

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-6+\frac{121}{16}$$

Add -6 to \frac{121}{16}.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{25}{16}$$

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}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{5}{4} x-\frac{11}{4}=-\frac{5}{4}$$

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

$$x=4 x=\frac{3}{2}$$