SOLUTION: Solve. (x + 3)^2 = 3 I came up with 5+-SQRT3 Did I just maybe get it right?

Algebra ->  Equations -> SOLUTION: Solve. (x + 3)^2 = 3 I came up with 5+-SQRT3 Did I just maybe get it right?      Log On


   



Question 84482: Solve. (x + 3)^2 = 3
I came up with 5+-SQRT3
Did I just maybe get it right?

Found 2 solutions by checkley75, rapaljer:
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
I'M AFRAID NOT!!!
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(X+3)^2=3
X^2+6X+9=3
X^2+6X+9-3=0
X^2+6X+6=0
USING THE QUDRATIC EQUATION:x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
X=(-6+-SQRT[6^2-4*6*1])/2*1
X=(-6+-SQRT[36-24])/2
X=(-6+-SQRT12)/2
X=(-6+-3.464)/2
X=(-6+3.464)/2
X=-2.536/2
X=-1.27 ANSWER.
X=(-6-3.464)/2
X=-9.464/2
X=-4.73 ANSWER.
PROOF
(-1.27+3)^2=3
1.73^2=3
3=3
(-4.73+3)^2=3
-1.73^2=3
3=3

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
This problem would have been a LOT easier to solve if, instead of using the quadratic formula as Checkley suggested you do, you just take the square root of each side:

%28x+%2B+3%29%5E2+=+3
%28x%2B3%29+=+0%2B-sqrt%283%29

Now, just subtract 3 from each side:
x%2B3-3=+-3%2B-sqrt%283%29+

You were ALMOST correct!! Never mind Checkley's chiding!!

R^2 at SCC