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Today we're going to deal with the problem.
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The longer leg of a right angle triangle is four feet longer than the shorter leg.
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We need to find the length of the two legs if the hypotenuse is 20 foot.
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So drawing this, we have a triangle where if we call the length of the shortest leg a, the length of the longest leg is a plus four, since it's four feet longer.
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And then we know if we call this b that a triangle, a right angle triangle obeys a squared plus b squared is equal to c squared, where the length of the hypotenies is c.
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But in this case, this is 20 squared, which is 400.
00:49
So now plugging everything in, we have a squared plus a plus 4 squared is equal to 400.
00:59
Expanding the bracket, we have a squared plus a squared plus 8a plus 16 is equal to 400.
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Now we can bring everything together.
01:14
We have 2a squared plus 8a plus 16 is equal to 400.
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Everything is even, so we can divide by a factor of 2.
01:23
So we get a squared plus 4a plus 8 is equal to 200.
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Now we can bring everything over to the left -hand side, and we get a squared plus 4a minus 192 is equal to 0.
01:43
Now, to solve this quadratic, we're looking to find two numbers, well, c and d, such that their product is minus 192, but their sum is 4.
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Now, because their product is a negative number, it's minus 192, we know that one of the numbers must be negative.
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So let's write down the factors of 192.
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We have 1 and 192, 2 and 96.
02:24
We have 3 times 64, will give us 192.
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And then it's also divisible by four...