SOLUTION: b^2+10^2=13^2 I have already tried these setups: 13^2=10^2+b^2 and 10^2=13^2+b^2

Algebra ->  Algebra  -> Pythagorean-theorem -> SOLUTION: b^2+10^2=13^2 I have already tried these setups: 13^2=10^2+b^2 and 10^2=13^2+b^2       Log On

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Question 120070This question is from textbook Algebra
: b^2+10^2=13^2
I have already tried these setups:
13^2=10^2+b^2 and
10^2=13^2+b^2
This question is from textbook Algebra

Answer by solver91311(12126) About Me  (Show Source):
You can put this solution on YOUR website!
You wrote:
"b%5E2%2B10%5E2=13%5E2

I have already tried these setups:

13%5E2=10%5E2%2Bb%5E2 and 10%5E2=13%5E2%2Bb%5E2"

10%5E2=13%5E2%2Bb%5E2 is incorrect: You have the signs wrong. In order to move the 10%5E2 and 13%5E2 terms to the opposite sides of the equal sign, you have to subtract them. The correct equation would be -10%5E2=-13%5E2%2Bb%5E2. Also neither of your arrangements, even though 13%5E2=10%5E2%2Bb%5E2 is technically correct, won't do you much good because you have not isolated the variable in either case.

The arrangement that you want comes from adding -10%5E2 to both sides of the original equation:

b%5E2%2B10%5E2=13%5E2

b%5E2%2B10%5E2-10%5E2=13%5E2-10%5E2

b%5E2=13%5E2-10%5E2

Notice that we now have the variable, b, isolated to one side of the equal sign and the constant values on the other side. Now you can take the square root of both sides:

sqrt%28b%5E2%29=sqrt%2813%5E2-10%5E2%29

b=sqrt%28169-100%29=sqrt%2869%29

So your solution set is b=sqrt%2869%29 or b=-sqrt%2869%29. However, if, as I suspect, you are dealing with a right triangle of hypotenuse 13 and long leg of 10 and want the measure of the short leg, you can exclude the b=-sqrt%2869%29 result because a negative number for a length is absurd. Also sqrt%2869%29 is in simplest terms because there are no perfect square factors of 69 (the prime factorization of 69 is 3 * 23).

Hope this helps.
John