SOLUTION: I need help in setting up an equation. The problem says the longer leg of a right triangle is 3 cm less than 5 times the shorter leg. The hypotenuse is 3 cm longer than the short

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: I need help in setting up an equation. The problem says the longer leg of a right triangle is 3 cm less than 5 times the shorter leg. The hypotenuse is 3 cm longer than the short      Log On


   

Question 689216: I need help in setting up an equation. The problem says the longer leg of a right triangle is 3 cm less than 5 times the shorter leg. The hypotenuse is 3 cm longer than the shorter leg. Find the length of the shorter leg. a^2 + b^2 = c^2 or in this case b^2 + (5b - 3)^2 = (5b + 3)^2. I am having problems once I get the equation to 26b^2 - 30B +9 = 25b^2 + 30B +9. Can someone help me from that point or tell me where I went wrong in setting the problem up?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If "the longer leg of a right triangle is 3 cm less than 5 times the shorter leg,"
you are correct in translating
a%5E2+%2B+b%5E2+=+c%5E2 into b%5E2+%2B+%285b+-+3%29%5E2+=+c%5E2, calling the shorter leg b.

However, if the problem says exactly that the "hypotenuse is 3 cm longer than the shorter leg," it should be
+b%5E2+%2B+%285b+-+3%29%5E2+=+%28b+%2B+3%29%5E2 instead of
b%5E2+%2B+%285b+-+3%29%5E2+=+%285b+%2B+3%29%5E2.
Then,
+b%5E2+%2B+%285b+-+3%29%5E2+=+%28b+%2B+3%29%5E2 --> 26b%5E2+-+30b+%2B9+=+b%5E2+%2B+6b+%2B9 --> 25b%5E2+-36b=0 --> 25b%28b-36%2F25%29=0
and as b=0 makes no triangle, it must be
highlight%28b=36%2F25%29cm, or, if you prefer highlight%28b=1.44%29cm.

You made the hypotenuse 3 cm longer than 5 times the shorter leg in
b%5E2+%2B+%285b+-+3%29%5E2+=+%285b+%2B+3%29%5E2.
If that was really what the problem stated, then
26b%5E2-30b%2B9=25b%5E2%2B30b%2B9 --> b%5E2-60b=0 --> b%28b-60%29=0
and the solution would be highlight%28b=60%29cm.