SOLUTION: I have a problem that I wanted to check to see if I've figured out the right answer. The hypotenuse of a right triangle is 4 inches long. one leg is 1 inch longer than the other. F

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Question 140230: I have a problem that I wanted to check to see if I've figured out the right answer. The hypotenuse of a right triangle is 4 inches long. one leg is 1 inch longer than the other. Find the length of the shorter leg to the nearest tenth.
I have 2.3 is this right. Want to make sure I'm understanding the problem correctly. Thank you.
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
4^2=x^2+(x+1)^2
16=x^2+x^2+2x+1
2x^2+2x+1-16=0
2x^2+2x-15=0
Using the quadratic equation:x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29we get:
x=(-2+-sqrt[-2^2-4*2*-15])/2*2
x=(-2+-sqrt[4+120])/4
x=(-2+-sqrt124)/4
x=(-2+-11.1355)/4
x=(-2+11.1355)/4
x=9.1355)/4
x=2.28 for the shorter leg.
2.28+1=3.28 for the longer leg.
proof
16=2.28^2+3.28^2
16=5.2+10.8
16=16