Question 391868
{{{(u-3)^2 = 2u^2 -3u -9}}}
I'm going to show you a clever way to solve this and a striaghtforward way to solve this.<br>
The clever way depends on recognizing that<ul><li>the equation is a quadratic equation. So you will want one side to be zero and the other side to be factored.</li><li>(u-3) is a factor of both sides of this equation</li></ul>
The straightforward way is to simplify both sides, get one side to be zero and then factor each side. This would be the only method if both sides did not have a common factor like (u-3).<br>
First the clever solution. We start by factoring the right side:
{{{(u-3)^2 = (2u + 3)(u-3)}}}
Then we subtract the entire right side from each side of the equation:
{{{(u-3)^2 - (2u + 3)(u-3) = 0}}}
Then we factor out the common factor, (u-3):
{{{(u-3)((u-3) - (2u+3)) = 0}}}
Simplifying the second factor we get:
{{{(u-3)(-u-6) = 0}}}
Now we use the Zero Product Property which tells us that this (or any) product can be zero <i>only</i> if one (or more) of the factors is zero. So:
u-3 = 0 or -u-6 = 0
Solving these we get:
u = 3 or u = -6<br>
{{{(u-3)^2 = 2u^2 -3u -9}}}
The straightforward way starts with simplifying both sides. Using FOIL or the {{{(a-b)^2 = a^2 -2ab + b^2}}} pattern to multiply out the left sides we get:
{{{(u)^2 - 2(u)(3) + (3)^2 = 2u^2 -3u -9}}}
which simplifies to:
{{{u^2 -6u + 9 = 2u^2 -3u -9}}}
Next we get one side to be zero. Since the step after this is factoring and since factoring is easier with a positive leading coefficient, I am going to subtract the entire left side from both sides:
{{{0 = u^2 +3u -18}}}
Factoring we get:
0 = (u-3)(u+6)
Using the Zero Product Property we get:
u-3 = 0 or u+6 = 0
Solving each of these we get:
u = 3 or u = -6