Solve Using the Quadratic Formula (3y-1)^2=0

${(3 y - 1)}^{2} = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 9$ , $b = - 6$ , and $c = 1$ into the quadratic formula and solve for $y$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (9 \cdot 1)}}{2 \cdot 9}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$y = \frac{6 \pm \sqrt{36 - 4 \cdot 9 \cdot 1}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot 1$ .

Multiply $- 4$ by $9$ .

$y = \frac{6 \pm \sqrt{36 - 36 \cdot 1}}{2 \cdot 9}$

Multiply $- 36$ by $1$ .

$y = \frac{6 \pm \sqrt{36 - 36}}{2 \cdot 9}$

Subtract $36$ from $36$ .

$y = \frac{6 \pm \sqrt{0}}{2 \cdot 9}$

Rewrite $0$ as $0^{2}$ .

$y = \frac{6 \pm \sqrt{0^{2}}}{2 \cdot 9}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \frac{6 \pm 0}{2 \cdot 9}$

$6$ plus or minus $0$ is $6$ .

$y = \frac{6}{2 \cdot 9}$

Multiply $2$ by $9$ .

$y = \frac{6}{18}$

Cancel the common factor of $6$ and $18$ .

Factor $6$ out of $6$ .

$y = \frac{6 (1)}{18}$

Cancel the common factors.

Factor $6$ out of $18$ .

$y = \frac{6 \cdot 1}{6 \cdot 3}$

Cancel the common factor.

$y = \frac{6 \cdot 1}{6 \cdot 3}$

Rewrite the expression.

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

Step 4

The final answer is the combination of both solutions.

$y = \frac{1}{3}$ Double roots