# How do you solve 8y^4 - 4y^2 = 0?

Jul 30, 2016

$y = 0 \text{ , } y = \pm \frac{\sqrt{2}}{2}$

#### Explanation:

Factor out ${y}^{2}$ giving:$\text{ } {y}^{2} \left(8 {y}^{2} - 4\right) = 0$

So ${y}^{2} = 0 \implies y = 0$

Or $\textcolor{b r o w n}{\left(8 {y}^{2} - 4\right) = 0} \textcolor{g r e e n}{\text{ " =>" " y^2=1/2)color(purple)(" " =>" } y = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}}$

Jul 30, 2016

$y = 0 \mathmr{and} y = \pm \sqrt{\frac{1}{2}}$

#### Explanation:

Find the factors first., look for a common factor before doing anything else.

$8 {y}^{4} - 4 {y}^{2} = 0$

$4 {y}^{2} \left(2 {y}^{2} - 1\right) = 0$

Either of the factors could be equal to 0.
Make two equations and solve each.

$\mathmr{if} 4 {y}^{2} = 0 \text{ "rArr y^2 = 0 " } \Rightarrow y = 0$

if $2 {y}^{2} - 1 = 0 \text{ "rArr2y^2 = 1" } \Rightarrow {y}^{2} = \frac{1}{2}$

$y = 0 \mathmr{and} y = \pm \sqrt{\frac{1}{2}}$

Jul 30, 2016

$y = - \frac{\sqrt{2}}{2} , \text{ "0," } \frac{\sqrt{2}}{2}$

#### Explanation:

First, notice that both terms have a common factor of $4 {y}^{2}$. This can be factored out of both terms.

$8 {y}^{4} - 4 {y}^{2} = 0$

$4 {y}^{2} \left(\frac{8 {y}^{4}}{4 {y}^{2}} - \frac{4 {y}^{2}}{4 {y}^{2}}\right) = 0$

$4 {y}^{2} \left(2 {y}^{2} - 1\right) = 0$

We now have two different terms being multiplied to equal $0$. The first is just $4 {y}^{2}$, and the other is the entirety of $\left(2 {y}^{2} - 1\right)$.

We can use the "Zero Factor Property" here, which basically states that if factors are being multiplied to equal $0$, it stands to reason that either factor must also equal $0$. (If $a b = 0$, then either $a = 0$ or b=0.)

So here, we know that:

$\left\{\begin{matrix}4 {y}^{2} = 0 \\ 2 {y}^{2} - 1 = 0\end{matrix}\right.$

Solving the first:

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

This boils down to our first solution:

$\overline{\underline{|}} \textcolor{w h i t e}{\frac{a}{a}} y = 0 \textcolor{w h i t e}{\frac{a}{a}} |$

Solving the second:

$2 {y}^{2} - 1 = 0$

$2 {y}^{2} = 1$

${y}^{2} = \frac{1}{2}$

Taking the square root of both sides (remember to allow positive and negative solutions):

$y = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$

Giving $2$ more solutions:

$\overline{\underline{|}} \textcolor{w h i t e}{\frac{a}{a}} y = \frac{\sqrt{2}}{2} \textcolor{w h i t e}{\frac{a}{a}} |$

$\overline{\underline{|}} \textcolor{w h i t e}{\frac{a}{a}} y = - \frac{\sqrt{2}}{2} \textcolor{w h i t e}{\frac{a}{a}} |$