SOLUTION: the question that i am asked is solve :{{{y^2+10y+25=9}}} so far i did was :{{{y^2+10y+25-9=0}}}, next i subtracted 25-9, i got {{{y^2+10y+16=0}}} then i got stuck so if you may

Algebra ->  Square-cubic-other-roots -> SOLUTION: the question that i am asked is solve :{{{y^2+10y+25=9}}} so far i did was :{{{y^2+10y+25-9=0}}}, next i subtracted 25-9, i got {{{y^2+10y+16=0}}} then i got stuck so if you may      Log On


   



Question 444596: the question that i am asked is solve :y%5E2%2B10y%2B25=9 so far i did was
:y%5E2%2B10y%2B25-9=0, next i subtracted 25-9, i got y%5E2%2B10y%2B16=0
then i got stuck so if you may please help me and explain it will be a great deal of help

Found 2 solutions by stanbon, oberobic:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
i got y%5E2%2B10y%2B16=0
then i got stuck so if you may please help me
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y^2+10y+16 = 0
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Use Inverse FOIL:
y^2+8y+2y+16 = 0
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y(y+8)+2(y+8) = 0
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(y+8)(y+2) = 0
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y = -8 or y = -2
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Cheers,
Stan H.
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Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
So far, so good.
But now you have to factor it.
What two terms can you multiply to get 16?
1 * 16
2 * 8
That's it.
...
Now add them together
1 + 16 = 17
2 + 8 = 10
That's the one you're looking for: 10, as in 10y.
.
So, we can factor the equation:
y%5E2+%2B+10y+%2B+16+=+%28y+%2B+2%29%28y+%2B+8%29
.
We can check this to be sure.
%28y+%2B+2%29%28y+%2B+8%29+=+y%5E2+%2B+8y+%2B+2y+%2B+16
y%5E2+%2B+8y+%2B+2y+%2B+16+=+y%5E2+%2B+10y+%2B+16
.
But really, we solve these types of equations to find values of y.
%28y+%2B+2%29%28y+%2B+8%29+=+0
So,
y = -2 or y = -8 to make the equation = 0.
.
graph%28500%2C500%2C-10%2C10%2C-10%2C10%2Cx%2Ax%2B10%2Ax%2B16%29