Solve 4x^2-4x+1=x^2-6x+9

$$4 x ^ { 2 } - 4 x + 1 = x ^ { 2 } - 6 x + 9$$

Solve for x

$$x=-2$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$4x^{2}-4x+1-x^{2}=-6x+9$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-4x+1=-6x+9$$

Add 6x to both sides.

$$3x^{2}-4x+1+6x=9$$

Combine -4x and 6x to get 2x.

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

Subtract 9 from both sides.

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

Subtract 9 from 1 to get -8.

$$3x^{2}+2x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=2 ab=3\left(-8\right)=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 2.

$$a=-4 b=6$$

Rewrite 3x^{2}+2x-8 as \left(3x^{2}-4x\right)+\left(6x-8\right).

$$\left(3x^{2}-4x\right)+\left(6x-8\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(3x-4\right)+2\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(x+2\right)$$

To find equation solutions, solve 3x-4=0 and x+2=0.

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

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$4x^{2}-4x+1-x^{2}=-6x+9$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-4x+1=-6x+9$$

Add 6x to both sides.

$$3x^{2}-4x+1+6x=9$$

Combine -4x and 6x to get 2x.

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

Subtract 9 from both sides.

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

Subtract 9 from 1 to get -8.

$$3x^{2}+2x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 3\left(-8\right)}}{2\times 3}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 3\left(-8\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-2±\sqrt{4-12\left(-8\right)}}{2\times 3}$$

Multiply -12 times -8.

$$x=\frac{-2±\sqrt{4+96}}{2\times 3}$$

Add 4 to 96.

$$x=\frac{-2±\sqrt{100}}{2\times 3}$$

Take the square root of 100.

$$x=\frac{-2±10}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-2±10}{6}$$

Now solve the equation x=\frac{-2±10}{6} when ± is plus. Add -2 to 10.

$$x=\frac{8}{6}$$

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-2±10}{6} when ± is minus. Subtract 10 from -2.

$$x=-\frac{12}{6}$$

Divide -12 by 6.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

Subtract x^{2} from both sides.

$$4x^{2}-4x+1-x^{2}=-6x+9$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-4x+1=-6x+9$$

Add 6x to both sides.

$$3x^{2}-4x+1+6x=9$$

Combine -4x and 6x to get 2x.

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

Subtract 1 from both sides.

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

Subtract 1 from 9 to get 8.

$$3x^{2}+2x=8$$

Divide both sides by 3.

$$\frac{3x^{2}+2x}{3}=\frac{8}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{2}{3}x=\frac{8}{3}$$

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.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{8}{3}+\left(\frac{1}{3}\right)^{2}$$

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

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{8}{3}+\frac{1}{9}$$

Add \frac{8}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{25}{9}$$

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}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{25}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{25}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{5}{3} x+\frac{1}{3}=-\frac{5}{3}$$

Subtract \frac{1}{3} from both sides of the equation.

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