SOLUTION: Segment JK=2x + 5 Segment KL=5x + 3 Segment JL=x^2 Find x. So far, we have x^2=2x+5+5x+3 x^2=7x+8 We don't know how to solve for x by showing our work. We ca

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Question 181440: Segment JK=2x + 5
Segment KL=5x + 3
Segment JL=x^2
Find x.
So far, we have x^2=2x+5+5x+3
x^2=7x+8
We don't know how to solve for x by showing our work. We can figure out that the answer is x=8, but can't work it out.
Thanks for any assistance!
Found 2 solutions by scott8148, stanbon:
Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website!
x^2=7x+8
subtracting 7x+8 .... x^2-7x-8 = 0
factoring .... (x+1)(x-8) = 0

x+1 = 0 .... x = -1 .... negative lengths not realistic

x-8 = 0 .... x = 8

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Segment JK=2x + 5
Segment KL=5x + 3
Segment JL=x^2
Find x.
So far, we have x^2=2x+5+5x+3
x^2=7x+8
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x^2 - 7x - 8 = 0
Factor to get:
(x-8)(x+1) = 0
x = 8 or x = -1
----------------------8:
For x = 8
JK = 2x+5 = 2*8 + 5 = 21
KL = 5x + 3 = 5*8+3 = 43
JL = x^2 = 8^2 = 64
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For x = -1
JK = 2x+5 = -2+5 = 3
KL = 5x+5 = 5*-1+5 = 0
JL = x^2 = (-1)^2 = 1
Note that 3+0 is not equal to 1
So x=-1 is not a true solution.
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Cheers,
Stan H.