Solve 6x-1=9x^2

$$6x-1=9 { x }^{ 2 }$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Factoring By Grouping

Subtract 9x^{2} from both sides.

$$6x-1-9x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-9x^{2}+6x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=6 ab=-9\left(-1\right)=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 6.

$$a=3 b=3$$

Rewrite -9x^{2}+6x-1 as \left(-9x^{2}+3x\right)+\left(3x-1\right).

$$\left(-9x^{2}+3x\right)+\left(3x-1\right)$$

Factor out -3x in -9x^{2}+3x.

$$-3x\left(3x-1\right)+3x-1$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(-3x+1\right)$$

To find equation solutions, solve 3x-1=0 and -3x+1=0.

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

Steps Using the Quadratic Formula

Subtract 9x^{2} from both sides.

$$6x-1-9x^{2}=0$$

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.

$$-9x^{2}+6x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 6 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-6±\sqrt{36+36\left(-1\right)}}{2\left(-9\right)}$$

Multiply 36 times -1.

$$x=\frac{-6±\sqrt{36-36}}{2\left(-9\right)}$$

Add 36 to -36.

$$x=\frac{-6±\sqrt{0}}{2\left(-9\right)}$$

Take the square root of 0.

$$x=-\frac{6}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=-\frac{6}{-18}$$

Reduce the fraction \frac{-6}{-18} to lowest terms by extracting and canceling out 6.

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

Steps for Completing the Square

Subtract 9x^{2} from both sides.

$$6x-1-9x^{2}=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$6x-9x^{2}=1$$

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.

$$-9x^{2}+6x=1$$

Divide both sides by -9.

$$\frac{-9x^{2}+6x}{-9}=\frac{1}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{6}{-9}x=\frac{1}{-9}$$

Reduce the fraction \frac{6}{-9} to lowest terms by extracting and canceling out 3.

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

Divide 1 by -9.

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

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=-\frac{1}{9}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

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

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

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=0$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{3}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{1}{3}=0 x-\frac{1}{3}=0$$

Add \frac{1}{3} to both sides of the equation.

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

The equation is now solved. Solutions are the same.

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