Solve x^2=x/9+5

$${ x }^{ 2 } = \frac{ x }{ 9+5 }$$

Solve for x

$$x=\frac{1}{14}\approx 0.071428571$$

$$x=0$$

Steps Using the Quadratic Formula

Add 9 and 5 to get 14.

$$x^{2}=\frac{x}{14}$$

Subtract \frac{x}{14} from both sides.

$$x^{2}-\frac{x}{14}=0$$

Multiply both sides of the equation by 14.

$$14x^{2}-x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 14 for a, -1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1}}{2\times 14}$$

Take the square root of 1.

$$x=\frac{-\left(-1\right)±1}{2\times 14}$$

The opposite of -1 is 1.

$$x=\frac{1±1}{2\times 14}$$

Multiply 2 times 14.

$$x=\frac{1±1}{28}$$

Now solve the equation x=\frac{1±1}{28} when ± is plus. Add 1 to 1.

$$x=\frac{2}{28}$$

Reduce the fraction \frac{2}{28} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{14}$$

Now solve the equation x=\frac{1±1}{28} when ± is minus. Subtract 1 from 1.

$$x=\frac{0}{28}$$

Divide 0 by 28.

$$x=0$$

The equation is now solved.

$$x=\frac{1}{14} x=0$$

Steps for Completing the Square

Add 9 and 5 to get 14.

$$x^{2}=\frac{x}{14}$$

Subtract \frac{x}{14} from both sides.

$$x^{2}-\frac{x}{14}=0$$

Multiply both sides of the equation by 14.

$$14x^{2}-x=0$$

Divide both sides by 14.

$$\frac{14x^{2}-x}{14}=\frac{0}{14}$$

Dividing by 14 undoes the multiplication by 14.

$$x^{2}-\frac{1}{14}x=\frac{0}{14}$$

Divide 0 by 14.

$$x^{2}-\frac{1}{14}x=0$$

Divide -\frac{1}{14}, the coefficient of the x term, by 2 to get -\frac{1}{28}. Then add the square of -\frac{1}{28} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{14}x+\left(-\frac{1}{28}\right)^{2}=\left(-\frac{1}{28}\right)^{2}$$

Square -\frac{1}{28} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{14}x+\frac{1}{784}=\frac{1}{784}$$

Factor x^{2}-\frac{1}{14}x+\frac{1}{784}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{28}\right)^{2}=\frac{1}{784}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{28}\right)^{2}}=\sqrt{\frac{1}{784}}$$

Simplify.

$$x-\frac{1}{28}=\frac{1}{28} x-\frac{1}{28}=-\frac{1}{28}$$

Add \frac{1}{28} to both sides of the equation.

$$x=\frac{1}{14} x=0$$