SOLUTION: i am having trouble solving this equation how can you factor 2a2 + 1 + 3a

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Question 622580: i am having trouble solving this equation
how can you factor
2a2 + 1 + 3a
Found 2 solutions by josmiceli, MathTherapy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+2a%5E2+%2B+1+%2B+2a+
+2a%5E2+%2B+2a+%2B+1+
Set it equal to zero
+2a%5E2+%2B+2a+%2B+1+=+0+
+2a%5E2+%2B+2a+=+-1+
+a%5E2+%2B+a+=+-1%2F2+
Complete the square
+a%5E2+%2B+a+%2B+%281%2F2%29%5E2+=+-1%2F2+%2B+%281%2F2%29%5E2+
+a%5E2+%2B+a+%2B+1%2F4++=+-2%2F4+%2B+1%2F4+
+%28+a+%2B+1%2F2%29%5E2+=+-1%2F4+
Take the square root of both sides
+a+%2B+1%2F2+=+%281%2F2%29%2Ai+
+a+=+%28+-1+%2B+i+%29%2F2+
and
+a+=+%28+-1+-+i+%29%2F2+
Hope I got it

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
i am having trouble solving this equation
how can you factor
2a2 + 1 + 3a

You DID NOT submit an equation, but a trinomial instead

2a%5E2++%2B+3a+%2B+1

You'll need to find two factors whose PRODUCT is "+ 2" (product of "c" and "a", and whose sum is + 3 (coefficient of b)

The two factors that satisfy this criteria are + 1, and + 2

We then have: highlight_green%28%282a+%2B+1%29%28a+%2B+1%29%29...That's it!!

If it's indeed an equation that you meant to post, then just set both binomials equal to 0 (zero), and solve for a to get 2 solutions.

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