# How do you solve  3 y ^2 - 48 = 0?

May 20, 2016

$y = \pm 4$

#### Explanation:

We see that there are only two terms in the given problem. One contains the unknown $y$ and other is a constant.
$3 {y}^{2} - 48 = 0$

Step 1. Keepning the term which contains $y$ on the LHS, move the contant term to RHS of the equation. We get

$3 {y}^{2} = 48$

Step 2. Divide both sides by the common facotor $3$.
$\frac{3 {y}^{2}}{3} = \frac{48}{3}$
Simplify

${y}^{2} = 16$

Step 3. To solve for $y$, Take square root of both sides
$\sqrt{{y}^{2}} = \sqrt{16}$
Simplify and remember to place $= \pm$ sign in front of one of the two terms. Selected RHS

$y = \pm 4$

May 20, 2016

$y = 4 , - 4$

#### Explanation:

$3 {y}^{2} - 48 = 0$

$3$ is a common factor between the two terms.
Take $3$ and write the remaining terms in brackets:

$3 \left({y}^{2} - 16\right) = 0$

Now, the product of $3$ and $\left({y}^{2} - 16\right)$ is $0$.

This means that $\textcolor{red}{\text{atleast ONE of the two terms is } 0}$.

$\implies 3 = 0 \text{ or } {y}^{2} - 16 = 0$

Obviously, $3 \ne 0$

Then, ${y}^{2} - 16 = 0$

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

This is in the form, $\textcolor{red}{{a}^{2} - {b}^{2}}$

And we know that, $\textcolor{red}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)}$

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

$\implies \left(y - 4\right) \left(y + 4\right) = 0$

Again, $\textcolor{red}{\text{atleast ONE of the two terms is } 0}$.

$\implies y - 4 = 0 \text{ or } y + 4 = 0$

$\implies y = 4 \text{ or } y = - 4$