SOLUTION: Determine the number and type of solutions for the equation 5x^2+40x+80=0

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Question 1132259: Determine the number and type of solutions for the equation 5x^2+40x+80=0
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
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+5x%5E2%2B40x%2B80=0+...simplify, both sides divide by 5
+x%5E2%2B8x%2B16=0+.......factor
+x%5E2%2B4x%2B4x%2B16=0+
+%28x%5E2%2B4x%29%2B%284x%2B16%29=0+
+x%28x%2B4%29%2B4%28x%2B4%29=0+
+%28x%2B4%29%28x%2B4%29=0+
solution: x=-4 multiplicity 2 or a double root

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Divide all the term by common factor 5 to get an equivalent equation in the simpler form


x^2 + 8x + 16 = 0.


Notice that it is


%28x%2B4%29%5E2 = 0.


The equation has single root x= -4.


It is the case, when two real solutions merge into one sigle value.


Answer.  The given equation has the single root x= -4, which is an integer number.