Solve x=9x^2-3x-1

$$x=9 { x }^{ 2 } -3x-1$$

Solve for x

$$x=\frac{\sqrt{13}+2}{9}\approx 0.622839031$$

$$x=\frac{2-\sqrt{13}}{9}\approx -0.178394586$$

Steps Using the Quadratic Formula

Subtract 9x^{2} from both sides.

$$x-9x^{2}=-3x-1$$

Add 3x to both sides.

$$x-9x^{2}+3x=-1$$

Combine x and 3x to get 4x.

$$4x-9x^{2}=-1$$

Add 1 to both sides.

$$4x-9x^{2}+1=0$$

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.

$$-9x^{2}+4x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 4 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-9\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-4±\sqrt{16+36}}{2\left(-9\right)}$$

Add 16 to 36.

$$x=\frac{-4±\sqrt{52}}{2\left(-9\right)}$$

Take the square root of 52.

$$x=\frac{-4±2\sqrt{13}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-4±2\sqrt{13}}{-18}$$

Now solve the equation x=\frac{-4±2\sqrt{13}}{-18} when ± is plus. Add -4 to 2\sqrt{13}.

$$x=\frac{2\sqrt{13}-4}{-18}$$

Divide -4+2\sqrt{13} by -18.

$$x=\frac{2-\sqrt{13}}{9}$$

Now solve the equation x=\frac{-4±2\sqrt{13}}{-18} when ± is minus. Subtract 2\sqrt{13} from -4.

$$x=\frac{-2\sqrt{13}-4}{-18}$$

Divide -4-2\sqrt{13} by -18.

$$x=\frac{\sqrt{13}+2}{9}$$

The equation is now solved.

$$x=\frac{2-\sqrt{13}}{9} x=\frac{\sqrt{13}+2}{9}$$

Steps for Completing the Square

Subtract 9x^{2} from both sides.

$$x-9x^{2}=-3x-1$$

Add 3x to both sides.

$$x-9x^{2}+3x=-1$$

Combine x and 3x to get 4x.

$$4x-9x^{2}=-1$$

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.

$$-9x^{2}+4x=-1$$

Divide both sides by -9.

$$\frac{-9x^{2}+4x}{-9}=-\frac{1}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{4}{-9}x=-\frac{1}{-9}$$

Divide 4 by -9.

$$x^{2}-\frac{4}{9}x=-\frac{1}{-9}$$

Divide -1 by -9.

$$x^{2}-\frac{4}{9}x=\frac{1}{9}$$

Divide -\frac{4}{9}, the coefficient of the x term, by 2 to get -\frac{2}{9}. Then add the square of -\frac{2}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{9}x+\left(-\frac{2}{9}\right)^{2}=\frac{1}{9}+\left(-\frac{2}{9}\right)^{2}$$

Square -\frac{2}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{9}x+\frac{4}{81}=\frac{1}{9}+\frac{4}{81}$$

Add \frac{1}{9} to \frac{4}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{9}x+\frac{4}{81}=\frac{13}{81}$$

Factor x^{2}-\frac{4}{9}x+\frac{4}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{9}\right)^{2}=\frac{13}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{9}\right)^{2}}=\sqrt{\frac{13}{81}}$$

Simplify.

$$x-\frac{2}{9}=\frac{\sqrt{13}}{9} x-\frac{2}{9}=-\frac{\sqrt{13}}{9}$$

Add \frac{2}{9} to both sides of the equation.

$$x=\frac{\sqrt{13}+2}{9} x=\frac{2-\sqrt{13}}{9}$$