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Question 838488: The measurement of two sides of a triangle are given {17x-7} and {3x^2+5}. The perimeter of the triangle is 13x^2-14X+12. Find the measurement of the third side.
I tried adding the two sides together and then subtracting from the perimeter but I don't know if the answer is correct. Just doesn't seem right to me.
3x^2+5+17x-7=3x^2+17x-2
perimeter 13x^2-14x+12
- total of the two sides (-)3x^2+17x+2
-3x^2-17x-2
my answer 10x^2-31x+10
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! Close but not quite right:
The sum of the two sides is 3x^2+17x-2. But somehow your subtracted 3x^2+17x+2 from the perimeter. So the constant term should be 12 - (-2) = 14, not 12 - 2 = 10.
P.S. In response to the questions in your "Thank you" note"- Once again, the sum of the two angles is 3x^2+17x-2 not 3x^2+17x+2
- Any time a subtraction is to be done, you have a choice of actually subtracting or adding the opposite. (For many reasons I highly recommend adding the opposite instead of subtracting.) So in this problem, you can either...
- Subtract:
(13x^2-14x+12) - (3x^2+17x-2)
which is difficult to get right; or... - Add the opposite:
(13x^2+(-14x)+12) + [-(3x^2+17x-2)]
which is the same as:
13x^2+(-14x)+12 + (-3x^2)+(-17x)+2
which is a lot easier to get right. (Note that the opposite of a polynomial has all of its signs changed.)
Either way, done correctly gives the answer of: 10x^2-31x+14
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