Solve (3x-1)^2=16

$$( 3 x - 1 ) ^ { 2 } = 16$$

Solve for x

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

$$x=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$9x^{2}-6x+1=16$$

Subtract 16 from both sides.

$$9x^{2}-6x+1-16=0$$

Subtract 16 from 1 to get -15.

$$9x^{2}-6x-15=0$$

Divide both sides by 3.

$$3x^{2}-2x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-2 ab=3\left(-5\right)=-15$$

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 -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

Rewrite 3x^{2}-2x-5 as \left(3x^{2}-5x\right)+\left(3x-5\right).

$$\left(3x^{2}-5x\right)+\left(3x-5\right)$$

Factor out x in 3x^{2}-5x.

$$x\left(3x-5\right)+3x-5$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(x+1\right)$$

To find equation solutions, solve 3x-5=0 and x+1=0.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$9x^{2}-6x+1=16$$

Subtract 16 from both sides.

$$9x^{2}-6x+1-16=0$$

Subtract 16 from 1 to get -15.

$$9x^{2}-6x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -6 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9\left(-15\right)}}{2\times 9}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 9\left(-15\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-6\right)±\sqrt{36-36\left(-15\right)}}{2\times 9}$$

Multiply -36 times -15.

$$x=\frac{-\left(-6\right)±\sqrt{36+540}}{2\times 9}$$

Add 36 to 540.

$$x=\frac{-\left(-6\right)±\sqrt{576}}{2\times 9}$$

Take the square root of 576.

$$x=\frac{-\left(-6\right)±24}{2\times 9}$$

The opposite of -6 is 6.

$$x=\frac{6±24}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{6±24}{18}$$

Now solve the equation x=\frac{6±24}{18} when ± is plus. Add 6 to 24.

$$x=\frac{30}{18}$$

Reduce the fraction \frac{30}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{6±24}{18} when ± is minus. Subtract 24 from 6.

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

Divide -18 by 18.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$9x^{2}-6x+1=16$$

Subtract 1 from both sides.

$$9x^{2}-6x=16-1$$

Subtract 1 from 16 to get 15.

$$9x^{2}-6x=15$$

Divide both sides by 9.

$$\frac{9x^{2}-6x}{9}=\frac{15}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{6}{9}\right)x=\frac{15}{9}$$

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

$$x^{2}-\frac{2}{3}x=\frac{15}{9}$$

Reduce the fraction \frac{15}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{2}{3}x=\frac{5}{3}$$

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.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=\frac{5}{3}+\left(-\frac{1}{3}\right)^{2}$$

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

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{5}{3}+\frac{1}{9}$$

Add \frac{5}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{16}{9}$$

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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

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

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

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