Solve (3x-3)^2=49

$$( 3 x - 3 ) ^ { 2 } = 49$$

Solve for x

$$x = \frac{10}{3} = 3\frac{1}{3} \approx 3.333333333$$

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

Steps Using Factoring By Grouping

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

$$9x^{2}-18x+9=49$$

Subtract 49 from both sides.

$$9x^{2}-18x+9-49=0$$

Subtract 49 from 9 to get -40.

$$9x^{2}-18x-40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=-18 ab=9\left(-40\right)=-360$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -360.

$$1,-360 2,-180 3,-120 4,-90 5,-72 6,-60 8,-45 9,-40 10,-36 12,-30 15,-24 18,-20$$

Calculate the sum for each pair.

$$1-360=-359 2-180=-178 3-120=-117 4-90=-86 5-72=-67 6-60=-54 8-45=-37 9-40=-31 10-36=-26 12-30=-18 15-24=-9 18-20=-2$$

The solution is the pair that gives sum -18.

$$a=-30 b=12$$

Rewrite 9x^{2}-18x-40 as \left(9x^{2}-30x\right)+\left(12x-40\right).

$$\left(9x^{2}-30x\right)+\left(12x-40\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(3x-10\right)+4\left(3x-10\right)$$

Factor out common term 3x-10 by using distributive property.

$$\left(3x-10\right)\left(3x+4\right)$$

To find equation solutions, solve 3x-10=0 and 3x+4=0.

$$x=\frac{10}{3} x=-\frac{4}{3}$$

Steps Using the Quadratic Formula

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

$$9x^{2}-18x+9=49$$

Subtract 49 from both sides.

$$9x^{2}-18x+9-49=0$$

Subtract 49 from 9 to get -40.

$$9x^{2}-18x-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -18 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 9\left(-40\right)}}{2\times 9}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 9\left(-40\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-18\right)±\sqrt{324-36\left(-40\right)}}{2\times 9}$$

Multiply -36 times -40.

$$x=\frac{-\left(-18\right)±\sqrt{324+1440}}{2\times 9}$$

Add 324 to 1440.

$$x=\frac{-\left(-18\right)±\sqrt{1764}}{2\times 9}$$

Take the square root of 1764.

$$x=\frac{-\left(-18\right)±42}{2\times 9}$$

The opposite of -18 is 18.

$$x=\frac{18±42}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{18±42}{18}$$

Now solve the equation x=\frac{18±42}{18} when ± is plus. Add 18 to 42.

$$x=\frac{60}{18}$$

Reduce the fraction \frac{60}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{10}{3}$$

Now solve the equation x=\frac{18±42}{18} when ± is minus. Subtract 42 from 18.

$$x=-\frac{24}{18}$$

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

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

The equation is now solved.

$$x=\frac{10}{3} x=-\frac{4}{3}$$

Steps for Completing the Square

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

$$9x^{2}-18x+9=49$$

Subtract 9 from both sides.

$$9x^{2}-18x=49-9$$

Subtract 9 from 49 to get 40.

$$9x^{2}-18x=40$$

Divide both sides by 9.

$$\frac{9x^{2}-18x}{9}=\frac{40}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{18}{9}\right)x=\frac{40}{9}$$

Divide -18 by 9.

$$x^{2}-2x=\frac{40}{9}$$

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

$$x^{2}-2x+1=\frac{40}{9}+1$$

Add \frac{40}{9} to 1.

$$x^{2}-2x+1=\frac{49}{9}$$

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

$$\left(x-1\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$x-1=\frac{7}{3} x-1=-\frac{7}{3}$$

Add 1 to both sides of the equation.

$$x=\frac{10}{3} x=-\frac{4}{3}$$