avai70178 one year ago Could someone help me solve this equation by completing the square? 6a^2-12a-93=-3

1. Astrophysics

To complete the square $x^2+ax=0$ we can do the following $x^2+\frac{ 2ax }{ 2 } + \left( \frac{ a }{ 2 } \right)^2-\left( \frac{ a }{ 2 } \right)^2=0$

2. Astrophysics

$6a^2-12a-90=0$ set it up as such now complete the square

3. avai70178

Wouldn't we add 93 to both sides so that we have $6a ^{2}-12a=90$ ?

4. avai70178

then divide -12 by two and square it and add that value to both sides?

5. avai70178

So that we have $6a ^{2}-12a+36=126$

6. avai70178

That's what I have so far

7. Astrophysics

Well I think it would be easiest if you first divide everything by 6 so we have $6a^2-12a-90=0 \implies a^2-2a-15=0$$\left( \frac{ -2 }{ 2 } \right)^2 = 1$ we do this to both sides, but you're essentially right it doesn't matter if you bring it to left or right side. So for simplicity in your case lets put it as $a^2-2a=15$ now we add the 1 to both sides so we get $a^2-2a+1=16$

8. Astrophysics

Now you can easily factor $a^2-2a+1$

9. avai70178

oh okay that way more simple thanks

10. Astrophysics

Yw :)