SOLUTION: i need help solving for a a^2+4a=45

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Question 67900: i need help solving for a
a^2+4a=45
Found 2 solutions by Earlsdon, Zoop:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for a:
a%5E2+%2B+4a+=+45 Subtract 45 from both sides.
a%5E2+%2B+4a+-+45+=+0 Solve this quadratic equation by factoring:
%28a-5%29%28a%2B9%29+=+0 Apply the zero product principle.
a-5+=+0 and/or a%2B9+=+0
If a-5+=+0 then a+=+5
If a%2B9+=+0 then a+=+-9
The solutions:
a = 5
a = -9

Answer by Zoop(21) About Me  (Show Source):
You can put this solution on YOUR website!
Equation: a%5E2%2B4a=45
Steps:
1. Make it a trinomial and move the 45 over.
a%5E2%2B4a-45
2. Factor it.
To factor this, find two factors of 45 that add to get 4a.
Those factors would be 9 and -5.
So, factored, it would be:
a%2B9%29%28a-5%29
3. Find the solutions.
When finding the solution of an equation like this, perform the same steps you use to find restrictions.
E.g. set each monomial to equal 0.
a%2B9=0 AND a-5=0
Solve each mini-equation.
The solutions are -9 and +5.
Try it out:
-9%5E2%2B4%28-9%29-45=0%3F
Yup.
How about:
5%5E2%2B4%285%29-45=0
Yup.
So your solution set is:
{-9,5}.