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
is incorrect: You have the signs wrong. In order to move the and terms to the opposite sides of the equal sign, you have to subtract them. The correct equation would be . Also neither of your arrangements, even though 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 to both sides of the original equation:
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:
So your solution set is or . 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 result because a negative number for a length is absurd. Also is in simplest terms because there are no perfect square factors of 69 (the prime factorization of 69 is 3 * 23).
