How do you solve $3 y^{2} - 12 = 0$ $3y^2-12=0$ ?

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How do you solve $3 y^{2} - 12 = 0$ $3y^2-12=0$ ?

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Answer 1 · 2018-03-13T23:11:22

$y = \pm 2$ $y = +- 2$

Explanation:

It is necessary to manipulate the equation so that the unknown variable (here, $y$ $y$ ) 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 $3 y^{2}$ $3 y^2$ by $3$ $3$ to remove the coefficient (and leave just $y^{2}$ $y^2$ ). It is necessary to do the same thing to both sides of the equation, so every term must be divided by $3$ $3$ .

That is,

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

implies

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

that is

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

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

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

To retrieve $y$ $y$ on its own, it is necessary to take the square root of both sides. As $4$ $4$ 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,

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

implies

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

that is

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

Answer 2 · 2018-03-13T23:16:21

$y = \pm 2$ $y = +-2$

Explanation:

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

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

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

$3 y (y + 2) - 6 (y + 2) = 0$ $3y(y+2)-6(y+2)=0$

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

$(3 y - 6) (y + 2) = 0$ $(3y-6) (y+2) = 0$

So

$3 y - 6 = 0 \Rightarrow y + 2 = 0$ $3y-6=0 => y+2=0$

$3 y = 6 \Rightarrow y = - 2$ $3y=6 => y=-2$

Answer 3 · 2018-03-13T23:42:15

$y = \pm 2$ $y = +-2$

Explanation:

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

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

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

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

$m m m m m m m m$ $color(white)(mmmmmmmm)$ ――――――――

Given $3 y^{2} - 12 = 0$ $3y^2−12=0$ Solve for $y$ $y$

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

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

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

$3 = 0$ $3= 0$ $\leftarrow$ $larr$ discarded solution

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

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

$m m m m m m m m$ $color(white)(mmmmmmmm)$ ――――――――

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