SOLUTION: Hi, I am having a little trouple with my home work. I would be vey grateful if some one could please help me with these two problems.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Hi, I am having a little trouple with my home work. I would be vey grateful if some one could please help me with these two problems.       Log On


   



Question 140885: Hi, I am having a little trouple with my home work. I would be vey grateful if some one could please help me with these two problems.
Solve using the square root property:
(x + 4)2 = 81
Solve by completing the square:
x2 + 6x + 2 = 0

Found 2 solutions by stanbon, rapaljer:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve using the square root property:
(x + 4)^2 = 81
Take the square root of each side to get:
x+4 = 9 or x+4 = -9
x = 5 or x = -13
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Solve by completing the square:
x^2 + 6x + 2 = 0
x^2 + 6x + 9 = -2 + 9
(x+3)^2 = 7
Take the square root to get:
x+3 = sqrt(7) or x+3 = -sqrt(7)
x = -3+sqrt(7) or x = -3-sqrt(7)
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Cheers,
Stan H.

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Solve using the square root property:
(x + 4)^2 = 81
This means taking the square root of both sides. When you do this, you have to remember to include the + or -:
%28x%2B4%29=0+%2B-+9
+x=-4+%2B-+9
x=-4%2B9 or x=-4-9
x=5 or x=-13

Solve by completing the square:
x%5E2+%2B+6x+%2B+2+=+0
x%5E2+%2B+6x++=+-2
x%5E2+%2B6x%2B9=-2%2B9
%28x%2B3%29%5E2=7
%28x%2B3%29=0%2B-+sqrt%287%29
+x=-3%2B-sqrt%287%29

R^2