SOLUTION: How can you solve a right triangle with two variables?? If a=x+1 b=x c=5 then how do you know?

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Question 1080879: How can you solve a right triangle with two variables?? If a=x+1 b=x c=5 then how do you know?
Found 3 solutions by Fombitz, MathLover1, Alan3354:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You only have one variable x.
I'm assuming that c is the hypotenuse.
a%5E2%2Bb%5E2=c%5E2
%28x%2B1%29%5E2%2Bx%5E2=25
x%5E2%2B2x%2B1%2Bx%5E2=25
2x%5E2%2B2x-24=0
x%5E2%2Bx-12=0
%28x%2B4%29%28x-3%29=0
Two solutions but only one is positive (which is required since we're getting a length).
x-3=0
x=3
So,
x%2B1=4
3-4-5 triangle

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

to solve a right triangle, use Pythagorean theorem c%5E2=a%5E2%2Bb%5E2

If a=x%2B1,+b=x and c=5, then we have:
5%5E2=%28x%2B1%29%5E2%2Bx%5E2
25=x%5E2%2B2x%2B1%2Bx%5E2
0=2x%5E2%2B2x%2B1-25
2x%5E2%2B2x-24=0.......simplify, divide by 2
x%5E2%2Bx-12=0....factor
x%5E2-3x%2B4x-12=0
%28x%5E2-3x%29%2B%284x-12%29=0
x%28x-3%29%2B4%28x-3%29=0
%28x+-+3%29+%28x+%2B+4%29+=+0
solutions:
%28x+-+3%29++=+0=>x=3
%28x+%2B4%29++=+0=>x=-4...........since side of the triangle cannot be negative, disregard x=-4
so,x=3
then
a=x%2B1->a=3%2B1->highlight%28a=4%29
+b=x ->highlight%28b=3%29
and highlight%28c=5%29

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
How can you solve a right triangle with two variables?? If a=x+1 b=x c=5 then how do you know?
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c is usually the hypotenuse.
Looks like the famous 3, 4, 5 right triangle.
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a^2 + b^2 = c^2
(x+1)^2 + x^2 = 25
x^2+2x+1 + x^2 = 25
2x^2 + 2x - 24 = 0
x^2 + x - 12 = 0
(x+4)*(x-3) = 0
x = 3
etc
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Ignore the x = -4