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+

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

Subtract 100 from both sides

Combine like terms

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

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=2, b=-12, and c=-80

Negate -12 to get 12

Square -12 to get 144 (note: remember when you square -12, you must square the negative as well. This is because .)

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 2 to get 4

So now the expression breaks down into two parts

or

Lets look at the first part:

Add the terms in the numerator
Divide

So one answer is

Now lets look at the second part:

Subtract the terms in the numerator
Divide

So another answer is

So our possible solutions are:
or

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

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

So the only solution is

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