# How do you solve 3y^2-12=0?

Mar 13, 2018

$y = \pm 2$

#### Explanation:

It is necessary to manipulate the equation so that the unknown variable (here, $y$) is on its own on the left hand side of the equation. It will be equal to whatever is on the right hand side of the equation.

To make a start, by inspection, it will be necessary to divide $3 {y}^{2}$ by $3$ to remove the coefficient (and leave just ${y}^{2}$). It is necessary to do the same thing to both sides of the equation, so every term must be divided by $3$.

That is,

$3 {y}^{2} - 12 = 0$

implies

$\frac{3 {y}^{2}}{3} - \frac{12}{3} = \frac{0}{3}$

that is

${y}^{2} - 4 = 0$

This may be rearranged (by adding $4$ to both sides of the equation) to yield

${y}^{2} = 4$

To retrieve $y$ on its own, it is necessary to take the square root of both sides. As $4$ is a perfect square, this will be easy but take care! Remember that square numbers have two roots, a positive one and a negative one.

So,

${y}^{2} = 4$

implies

$\sqrt{{y}^{2}} = \sqrt{4}$

that is

$y = 2$
or
$y = - 2$

Mar 13, 2018

$y = \pm 2$

#### Explanation:

$3 {y}^{2} - 12 = 0$

Multiply $3 \times \left(- 12\right)$ to get $- 36$ and use that to find two factors that when multiplied give $- 36$ and when added give $0$

$3 {y}^{2} + 6 y - 6 y - 12 = 0$

$3 y \left(y + 2\right) - 6 \left(y + 2\right) = 0$

Pull out the factors of the equation $3 {y}^{2} - 12 = 0$

$\left(3 y - 6\right) \left(y + 2\right) = 0$

So

$3 y - 6 = 0 \implies y + 2 = 0$

$3 y = 6 \implies y = - 2$

Mar 13, 2018

$y = \pm 2$

#### Explanation:

This is a Difference of Two Squares problem -- with a slight disguise because it is multiplied by $3$

The Difference of Two Squares is a case of Special Factoring.

By memorization, the Difference of Two Squares factors like this:

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

$\textcolor{w h i t e}{m m m m m m m m}$――――――――

Given   3y^2−12=0    Solve for $y$

1) Factor out the 3 to see the perfect squares
3 (y^2 - 4) = 0

2) Factor the Difference of the Two Squares
3 (y+2)(y-2)= 0

3) Set the factors equal to zero and solve for $y$

$3 = 0$ $\leftarrow$ discarded solution

$y + 2 = 0$
$y = - 2$ $\leftarrow$ one answer

$y - 2 = 0$
$y = 2$ $\leftarrow$ the other answer

$\textcolor{w h i t e}{m m m m m m m m}$――――――――

Here's a video you can watch to see more about Special Factoring