avai70178
  • avai70178
Could someone help me solve this equation by completing the square? 6a^2-12a-93=-3
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Astrophysics
  • Astrophysics
To complete the square \[x^2+ax=0\] we can do the following \[x^2+\frac{ 2ax }{ 2 } + \left( \frac{ a }{ 2 } \right)^2-\left( \frac{ a }{ 2 } \right)^2=0\]
Astrophysics
  • Astrophysics
\[6a^2-12a-90=0\] set it up as such now complete the square
avai70178
  • avai70178
Wouldn't we add 93 to both sides so that we have \[6a ^{2}-12a=90\] ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

avai70178
  • avai70178
then divide -12 by two and square it and add that value to both sides?
avai70178
  • avai70178
So that we have \[6a ^{2}-12a+36=126\]
avai70178
  • avai70178
That's what I have so far
Astrophysics
  • 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 \[a^2-2a+1=16\]
Astrophysics
  • Astrophysics
Now you can easily factor \[a^2-2a+1\]
avai70178
  • avai70178
oh okay that way more simple thanks
Astrophysics
  • Astrophysics
Yw :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.