Question

completing the square

Answer 1

\[4y^2-y=12\]

Answer 2

\[
(a+b)^2 = a^2+2ab+b^2
\]In this case \(a^2 = 4y^2\) and \(2ab = -y\).
So find \(a\). Then find \(b\) and \(b^2\).

Answer 3

@mendeaar002 Can you do that?

Answer 4

I'm kind of confused though could you write in steps

Answer 5

\[
(a+b)^2
\]Is the "square". We when you multiply it out you get: \[
(a+b)^2 = (a+b)(a+b) =a\cdot a+b\cdot a+b\cdot a+b\cdot b = a^2+2ab+b^2
\]

Answer 6

\[
4y^2-y=12\implies 4y^2-y-12=0
\]is in it's expanded form. We want to reverse this to get it into its square form.

Answer 7

\[
\begin{array}{cccc}

a^2&+&2ab&+&b^2 \\
(4y^2)&+&(−y)&+&(−12)
\end{array}
\]See how they correspond?

Answer 8

However it might not be the case that \(
4y^2−y−12
\) is a perfect square... we might need to add stuff to it to make it into a perfect square.

Answer 9

So our first job is to identify \(a\) and \(b\) given that: \[
\begin{array}{rcl}
a^2 &=& 4y^2 \\
2ab &=& -y
\end{array}
\]

Answer 10

Finding \(a\) is easy. Just take the square root on both sides of the first equation.

Answer 11

\[
a = 2y
\]

Answer 12

Then we can plug it into the second equation to find \(b\) \[
\begin{array}{rcl}
2(2y)b &=& -y \\
4yb &=& -y \\
b &=& -1y/4y &=&-1/4 \\
\end{array}
\]

Answer 13

Next we find \(b^2\): \[
b^2 = (-1/4)^2 = 1/16
\]

Answer 14

We have to pull \(b^2\) from our \(-12\).