Solve 2(3x-4)^2+1=9

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

Solve for x

$$x=2$$

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-4\right)^{2}.

$$2\left(9x^{2}-24x+16\right)+1=9$$

Use the distributive property to multiply 2 by 9x^{2}-24x+16.

$$18x^{2}-48x+32+1=9$$

Add 32 and 1 to get 33.

$$18x^{2}-48x+33=9$$

Subtract 9 from both sides.

$$18x^{2}-48x+33-9=0$$

Subtract 9 from 33 to get 24.

$$18x^{2}-48x+24=0$$

Divide both sides by 6.

$$3x^{2}-8x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-8 ab=3\times 4=12$$

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 12.

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

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

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

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

Factor out 3x in the first and -2 in the second group.

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

Factor out common term x-2 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-4\right)^{2}.

$$2\left(9x^{2}-24x+16\right)+1=9$$

Use the distributive property to multiply 2 by 9x^{2}-24x+16.

$$18x^{2}-48x+32+1=9$$

Add 32 and 1 to get 33.

$$18x^{2}-48x+33=9$$

Subtract 9 from both sides.

$$18x^{2}-48x+33-9=0$$

Subtract 9 from 33 to get 24.

$$18x^{2}-48x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, -48 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-48\right)±\sqrt{\left(-48\right)^{2}-4\times 18\times 24}}{2\times 18}$$

Square -48.

$$x=\frac{-\left(-48\right)±\sqrt{2304-4\times 18\times 24}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-48\right)±\sqrt{2304-72\times 24}}{2\times 18}$$

Multiply -72 times 24.

$$x=\frac{-\left(-48\right)±\sqrt{2304-1728}}{2\times 18}$$

Add 2304 to -1728.

$$x=\frac{-\left(-48\right)±\sqrt{576}}{2\times 18}$$

Take the square root of 576.

$$x=\frac{-\left(-48\right)±24}{2\times 18}$$

The opposite of -48 is 48.

$$x=\frac{48±24}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{48±24}{36}$$

Now solve the equation x=\frac{48±24}{36} when ± is plus. Add 48 to 24.

$$x=\frac{72}{36}$$

Divide 72 by 36.

$$x=2$$

Now solve the equation x=\frac{48±24}{36} when ± is minus. Subtract 24 from 48.

$$x=\frac{24}{36}$$

Reduce the fraction \frac{24}{36} to lowest terms by extracting and canceling out 12.

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

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-4\right)^{2}.

$$2\left(9x^{2}-24x+16\right)+1=9$$

Use the distributive property to multiply 2 by 9x^{2}-24x+16.

$$18x^{2}-48x+32+1=9$$

Add 32 and 1 to get 33.

$$18x^{2}-48x+33=9$$

Subtract 33 from both sides.

$$18x^{2}-48x=9-33$$

Subtract 33 from 9 to get -24.

$$18x^{2}-48x=-24$$

Divide both sides by 18.

$$\frac{18x^{2}-48x}{18}=-\frac{24}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x^{2}+\left(-\frac{48}{18}\right)x=-\frac{24}{18}$$

Reduce the fraction \frac{-48}{18} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{8}{3}x=-\frac{24}{18}$$

Reduce the fraction \frac{-24}{18} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{8}{3}x=-\frac{4}{3}$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=-\frac{4}{3}+\frac{16}{9}$$

Add -\frac{4}{3} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{4}{9}$$

Factor x^{2}-\frac{8}{3}x+\frac{16}{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}.

$$\left(x-\frac{4}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

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

Add \frac{4}{3} to both sides of the equation.

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