SOLUTION: I'm sorry. I know this problem is very similar to the example, but I get a very crazy answer when I've tried to do this like the example many times. I must be doing something wro

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Question 193575: I'm sorry. I know this problem is very similar to the example, but I get a very crazy answer when I've tried to do this like the example many times. I must be doing something wrong. Please help.
One leg of a right triangle is 96 inches. Find the hypotenuse and teh other leg if the length of the hypotenuse exceeds 2 and 1/2 times teh other leg by 4 inches.
Found 2 solutions by nerdybill, stanbon:
Answer by nerdybill(7384)   (Show Source): You can put this solution on YOUR website!
One leg of a right triangle is 96 inches. Find the hypotenuse and teh other leg if the length of the hypotenuse exceeds 2 and 1/2 times teh other leg by 4 inches.
.
Let x = length of "other leg" of triangle
then
(5/2)x-4 = hypotenuse
.
The applying the Pythagorean theorem:
x^2 + ((5/2)x-4) = 96^2
x^2 + ((5/2)x-4)((5/2)x-4) = 96^2
x^2 + (25/4)x^2-20x +16 = 96^2
(29/4)x^2-20x +16 = 9216
(29/4)x^2-20x -9200 = 0
29x^2 - 80x -36800 = 0
Solving using the quadratic equation we get:
x ={37.03, -34.26}
Throwing away the negative solution we're left with:
x = 37.03 inches (other leg)
.
Hypotenuse:
(5/2)x-4 = (5/2)37.03-4 = 88.58 inches (hypotenuse)
.
Details of the quadratic equation:
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=4275200 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 37.028546160883, -34.2699254712278. Here's your graph:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
One leg of a right triangle is 96 inches. Find the hypotenuse and the
other leg if the length of the hypotenuse exceeds 2 and 1/2 times teh other leg by 4 inches.
--------------------
One leg is 96 in
Let the "other leg" be "x" inches.
hypotenuse = (5/2)x + 4
----------------------------
Equation:
h^2 = leg^2 + "otherleg"^2
[(5/2)x + 4]^2 = 96^2 + x^2
[(5x+ 8)/2^2 = 96^2 + x^2
(5x+8)^2 = 4*96^2 + 4x^2
25x^2 + 90x + 91 = 36864 + 4x^2
21x^2 + 90x - 36773 = 0
-----
Positive solution:
x = 39.76
hypotenuse = (5/2)x+4 = 103.4
====================================
Cheers,
Stan H.

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