Help with a pythagorean theorom problem.

O

ochocki

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Feb 7, 2005
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Find the length of the shorter leg of a right triangle if the longer leg is 24 meters qnd the hypotenuse is 6 more than twice the shorter leg.

I set it up like this:

24² + x² = (2x + 6)²

Now I just can't quite figure out where to go from here, seeing that the 2x will be squared to 4x², and that there is an equal sign, I believe it could be set to zero. Can anyone give me some direction as to working this problem?
 
\(\displaystyle \L\begin{array}{rcl}
24^2 + x^2 & = & (2x + 6)^2 \\
& = & 4x^2 + 24x + 36 \\
\end{array}.\)

Now, just solve!
 
Man, I just can't seem to catch on to this.

4x² + 24x +36

I can factor out a 2 to get

2(2x²+12x+18)

But that doesn't seem right. Also I can't find a product of 4 & 36 that adds up to 24 so I can group it.

I think I am missing something very obvious here, and how did 24² + x² turn into 24x?
 
You must solve this:
\(\displaystyle \L
3x^2 + 24x - 540 = 0\)
 
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