Solve 9x^2+1=6x

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

Solve for x

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

Steps Using Factoring By Grouping

Subtract 6x from both sides.

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

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.

$$a+b=-6 ab=9\times 1=9$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. 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 the first and -1 in the second group.

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

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

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

Rewrite as a binomial square.

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

To find equation solution, solve 3x-1=0.

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

Steps Using the Quadratic Formula

Subtract 6x from both sides.

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

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}.

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

Square -6.

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

Multiply -4 times 9.

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

Add 36 to -36.

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

Take the square root of 0.

$$x=-\frac{-6}{2\times 9}$$

The opposite of -6 is 6.

$$x=\frac{6}{2\times 9}$$

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 6x from both sides.

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 9.

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

Dividing by 9 undoes the multiplication by 9.

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

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 -\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.

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