SOLUTION: Could I have some help figuring out this problem. Thank you. The hypotenuse of a right triagle is 4 inches long. One leg is 1 inch longer than the other. Find the length of the sho

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Question 190583: Could I have some help figuring out this problem. Thank you. The hypotenuse of a right triagle is 4 inches long. One leg is 1 inch longer than the other. Find the length of the shorter leg. Round to the nearest tenth.
Answer by nerdybill(7384) (Show Source):
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Could I have some help figuring out this problem. Thank you. The hypotenuse of a right triagle is 4 inches long. One leg is 1 inch longer than the other. Find the length of the shorter leg. Round to the nearest tenth.
.
Let x = length of shorter leg
then
x+1 = length of longer leg
.
Applying Pythagorean theorem we have:
x^2 + (x+1)^2 = 4^2
x^2 + x+^2+2x+1 = 16
2x+^2+2x+1 = 16
2x+^2+2x+-15 = 0
.
Applying the quadratic equation yields two solution:
x = {2.3, -3.3}
We can toss out the negative solution leaving us with:
x = 2.3 inches
.
Details of the quadratic follows:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=124 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 2.28388218141501, -3.28388218141501. Here's your graph: