SOLUTION: How many real solutions does the following problem have? Log(4x^2+4x+101)=2

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Question 375426: How many real solutions does the following problem have?
Log(4x^2+4x+101)=2
Found 2 solutions by Fombitz, ewatrrr:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
log%28%284x%5E2%2B4x%2B101%29%29=2
4x%5E2%2B4x%2B101=10%5E2
4x%5E2%2B4x%2B1=0
%282x%2B1%29%5E2=0
2x%2B1=0
2x=-1
highlight%28x=-1%2F2%29
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.
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1 solution, a double root at x=-1%2F2

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

Log(4x^2+4x+101)=2



10^2 = 4x^2 + 4x + 101

100 = 4x^2 + 4x + 101

0 = 4x^2 + 4x + 1

factoring

(2x + 1)(2x + 1)= 0 Note: the SUM of inner product(2x) and outer product(2x) = 4x

2x + 1 = 0

x = (-1/2)

there is only one real solution