SOLUTION: The hypotenuse of a right triangle is twice as long as one of the legs and 10 inches longer than the other. What are the lengths of the sides of the triangle? So far I know that

Question 156850: The hypotenuse of a right triangle is twice as long as one of the legs and 10 inches longer than the other. What are the lengths of the sides of the triangle?
So far I know that i should use the Pythagorean theorem.
I've drawn the triangle and know:
1st Leg: h/2
2nd Leg: h-10
Hypotenuse: h
(h-10)^2 + (h/2)^2 = h^2
4(h^2-20h-100+ (h^2/4))=h^2
4h^2-80h-400+h^2=h^2
5h^2-80h-400=h^2

Found 2 solutions by jim_thompson5910, nerdybill:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
You have the right equations set up, but your work is a little off.

Start with the given equation

FOIL. Note: the third term is +100 (not -100)

Square to get

Multiply EVERY term (including the terms on the right side) by the LCD 4 to eliminate the fraction

Subtract from both sides

Combine like terms.

Notice we have a quadratic equation in the form of where , , and

Let's use the quadratic formula to solve for h

Start with the quadratic formula

Plug in , , and

Negate to get .

Square to get .

Multiply to get

Subtract from to get

Multiply and to get .

Simplify the square root (note: If you need help with simplifying square roots, check out this solver)

Break up the fraction.

Reduce.

or Break up the expression.

or Now approximate the values of "h"

So the possible hypotenuses are or

However, if you plug into , you'll get a negative answer. So the only solution is

So the length of the hypotenuse is approximately units

So the first leg is units long and the second leg is units long
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
The hypotenuse of a right triangle is twice as long as one of the legs and 10 inches longer than the other. What are the lengths of the sides of the triangle?
.
So far I know that i should use the Pythagorean theorem.
I've drawn the triangle and know:
1st Leg: h/2
2nd Leg: h-10
Hypotenuse: h
.
(h-10)^2 + (h/2)^2 = h^2 <<-- formula looks good!
4(h^2-20h-100+ (h^2/4))=h^2 <<--why didn't you multiply the right side by 4?
.
Taking it from the top:
(h-10)^2 + (h/2)^2 = h^2
(h^2-20h+100) + (h^2/4) = h^2
4(h^2-20h+100) + 4(h^2/4) = 4h^2
4(h^2-20h+100) + h^2 = 4h^2
4h^2-80h+400 + h^2 = 4h^2
5h^2-80h+400 = 4h^2
h^2-80h+400 = 0
.
Can't factor so you must use the quadratic equation.
Doing so results in:
h = {74.64, 5.36}
If we pick h=5.36, our 2nd leg would be negative -- throw out solution.
Conclusion:
h = 74.64
1st Leg: h/2 = 37.32 inches
2nd Leg: h-10 = 64.64 inches
.
Reference: calculations for the quadratic

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=4800 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 74.6410161513776, 5.35898384862245. Here's your graph:

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