1. ## Quadratic Equat & Functions

My problem I have is on shapes which I get really confused on in story type problems and trying to understand how to get the quadratic equation to be able to fit the story and to solve it?

The problem:

The hypotenuse of a right triangle is 6 feet long. One leg is 2 feet shorter than the other. Find the lengths of the legs and round to the nearest tenth of a foot.

2. Originally Posted by kbryant05
...
The problem:
The hypotenuse of a right triangle is 6 feet long. One leg is 2 feet shorter than the other. Find the lengths of the legs and round to the nearest tenth of a foot.
Hello,

let x be the longer of the two legs, then the shorter leg is (x-2). (All mesures in feet).
By the rule of Pythagoras you can calculate:

$x^2+(x-2)^2=6^2$

Expand the LHS of your equation (be aware that you have to use the binomial formula!) and subtract 36 on both sides. You'll get:

$2x^2-4x-32=0$

Use the formula to solve this equation. The negative solution isn't very plausible with your problem. So you have:
$x=1+\sqrt{17}\ \approx\ 5.1'$

Greetings

EB

3. ## Can you show the second part?

thank you for being so very quick ... are you able to show me the second part of the equation for the other leg? They have 3.1 feet as an answer and I am not getting it? I'm a little lost. Sorry I'm just a farm girl who isn't very good with these types of problems.

4. Originally Posted by kbryant05
thank you for being so very quick ... are you able to show me the second part of the equation for the other leg? They have 3.1 feet as an answer and I am not getting it? I'm a little lost. Sorry I'm just a farm girl who isn't very good with these types of problems.
Hello,

as I've posted above the longer leg is 5.1', so the shorter leg must be 5.1' - 2' = 3.1'.

Greetings

EB

5. ## Sorry

I'm so sorry I was working on another problem at the same time and should have paid better attention. Thank you for such a fine job!!