SOLUTION: Hi, I was wondering if someone could possibly help me. I am supposed to do these homework questions in my Algebra class, but I am having a hard time getting started. I don't know w

Question 361348: Hi, I was wondering if someone could possibly help me. I am supposed to do these homework questions in my Algebra class, but I am having a hard time getting started. I don't know what steps to go through to obtain the answer. The back of the book shows the answers, but that doesn't help me much. I need to show I got to the answer. If anyone could possibly assist me, by showing me how you obtained the answer, I would be forever grateful! :o)
Section 11.2, Pg. 764, #79
The longer leg of a right triangle exceeds the shorter leg by 1 inch, and the hypotenuse exceeds the longer leg by 7 inches. Find the lengths of the legs. Round to the nearest tenth of a inch.
The answer the book shows is: 17.6in. and 18.6 in.
I don't understand how they were able to get this answer. Any help is greatly appreciated. Thanks!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The longer leg of a right triangle exceeds the shorter leg by 1 inch, and the hypotenuse exceeds the longer leg by 7 inches. Find the lengths of the legs. Round to the nearest tenth of a inch.
The answer the book shows is: 17.6in. and 18.6 in.
----
shorter leg: x
longer leg: x+1
hypotenuse: x+8
(x+8)^2 = (x+1)^2 + x^2
----
x^2 + 16x + 64 = x^2+2x+1 + x^2
x^2 -14x -63 = 0
-----
Quadratic formula:
x = [-b +- sqrt(b^2-4ac)]/(2a)
----
x = [14 +- sqrt(196-4*1*-63)/(2)
-----
x = [14 +- sqrt(448)]/2
---
x = [14 +- 21.17]/2
Positive solution:
x = 17.6
x+1 = 18.6
----
shorter leg: x = 17.6
longer leg: x+1 = 18.6
---------------------------
Cheers,
Stan H.
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