Question 390357: Find 3 consecutive integers such that the square of the first, added to the last, is 8
Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=first integer; x^2=square of the first
x+1=second
x+2=third
Now we are told that:
x^2+x+2=8 subtract 8 from each side
x^2+x-6=0 quadratic in standard form and it can be factored
(x+3)(x-2)=0
x=-3
and
x=2
First set of solutions
x=-3
x+1=-2
x+2=-1
CK
square of first is 9
9+(-1)=8
8=8
Second set of solutions
x=2
x+1=3
x+2=4
CK
square of first=4
4+4=8
Hope this helps--ptaylor
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