SOLUTION: I need help solving this problem. this is the work i did so far: there is a triangle on one side there is (x-2) the other side is (x-4) and the longest side is 10. (x-2)^2+

Algebra ->  Pythagorean-theorem -> SOLUTION: I need help solving this problem. this is the work i did so far: there is a triangle on one side there is (x-2) the other side is (x-4) and the longest side is 10. (x-2)^2+       Log On


   

Question 92081: I need help solving this problem. this is the work i did so far:
there is a triangle on one side there is (x-2) the other side is (x-4) and the longest side is 10.
(x-2)^2+ (x-4)^2= 10^2
(x-2)(x-2)+ (x-4)(x-4)= 100
x^2 + -2x + -2x + 4 + x^2 + -4x + -4x + 16
2x^2 + -12x + 20 = 100

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll pick up where you left off (your steps are correct)

2x%5E2+-12x+%2B+20+=+100+

2x%5E2+-12x+%2B+20+-+100=0+ Subtract 100 from both sides

2x%5E2+-12x+-80=0+ Combine like terms


Now let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29

So lets solve 2%2Ax%5E2-12%2Ax-80=0 ( notice a=2, b=-12, and c=-80)

x+=+%28--12+%2B-+sqrt%28+%28-12%29%5E2-4%2A2%2A-80+%29%29%2F%282%2A2%29 Plug in a=2, b=-12, and c=-80



x+=+%2812+%2B-+sqrt%28+%28-12%29%5E2-4%2A2%2A-80+%29%29%2F%282%2A2%29 Negate -12 to get 12



x+=+%2812+%2B-+sqrt%28+144-4%2A2%2A-80+%29%29%2F%282%2A2%29 Square -12 to get 144 (note: remember when you square -12, you must square the negative as well. This is because %28-12%29%5E2=-12%2A-12=144.)



x+=+%2812+%2B-+sqrt%28+144%2B640+%29%29%2F%282%2A2%29 Multiply -4%2A-80%2A2 to get 640



x+=+%2812+%2B-+sqrt%28+784+%29%29%2F%282%2A2%29 Combine like terms in the radicand (everything under the square root)



x+=+%2812+%2B-+28%29%2F%282%2A2%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%2812+%2B-+28%29%2F4 Multiply 2 and 2 to get 4

So now the expression breaks down into two parts

x+=+%2812+%2B+28%29%2F4 or x+=+%2812+-+28%29%2F4

Lets look at the first part:

x=%2812+%2B+28%29%2F4

x=40%2F4 Add the terms in the numerator
x=10 Divide

So one answer is
x=10



Now lets look at the second part:

x=%2812+-+28%29%2F4

x=-16%2F4 Subtract the terms in the numerator
x=-4 Divide

So another answer is
x=-4

So our possible solutions are:
x=10 or x=-4





Now lets find each of the leg's lengths:
Leg A:
10-2=8 Plug in x=10
-4-2=-6 Plug in x=-4
Since a negative length doesn't make sense, the solution x=-4 must be discarded


Leg B:
10-4=6 Plug in x=10
-4-4=-8 Plug in x=-4
Since a negative length doesn't make sense, the solution x=-4 must be discarded


So the only solution is

x=10 where the lengths of the legs are 8 and 6 (or 6 and 8)