SOLUTION: one leg of a right triangle exceeds the other leg by four inches. the hypotenuse is 10 inches. find the length of the shorter leg of the right triangle .

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: one leg of a right triangle exceeds the other leg by four inches. the hypotenuse is 10 inches. find the length of the shorter leg of the right triangle .      Log On


   

Question 270407: one leg of a right triangle exceeds the other leg by four inches. the hypotenuse is 10 inches. find the length of the shorter leg of the right triangle .
Answer by Greenfinch(383) About Me  (Show Source):
You can put this solution on YOUR website!
By Pythagoras
x^2 + (x + 4)^2 = 100
2x^2 + 8x + 16 = 100
x^2 + 4x - 42 = 0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B4x%2B-42+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%284%29%5E2-4%2A1%2A-42=184.

Discriminant d=184 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+184+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%284%29%2Bsqrt%28+184+%29%29%2F2%5C1+=+4.78232998312527
x%5B2%5D+=+%28-%284%29-sqrt%28+184+%29%29%2F2%5C1+=+-8.78232998312527

Quadratic expression 1x%5E2%2B4x%2B-42 can be factored:
1x%5E2%2B4x%2B-42+=+1%28x-4.78232998312527%29%2A%28x--8.78232998312527%29
Again, the answer is: 4.78232998312527, -8.78232998312527. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B-42+%29