Triangle JKL is a right triangle. JK=3x-6 KL=2x+11 JL=20. Find all the values of x. I know there are supposed to be three solutions and that you are supposed to use the quadratic formula but I have no idea how to use it or how to apply it to this problem. Please help!
You're told that JKL is a right triangle, but you don't know which side is the hypotenuse. For the moment, suppose that JL is the hypotenuse. Then set it up with the Pythagorean Theorem.
Expand all the squares and bring all the terms to one side. You can now apply the quadratic formula to solve for x.
Recall that you are dealing with sides of a triangle. These side lengths must be positive: 3x-6>0 and 2x+11>0. These will impose further restrictions on the solutions you get to the quadratic equation. (There is also the triangle inequality--that the sum of two side lengths exceed the other side's length--but the positivity restrictions will probably be sufficient.)
At the beginning, it was supposed that JL was the hypotenuse. This is one of three possible scenarios. Repeat the procedure supposing JK is the hypotenuse, then again with KL.
Thanks for asking,