solve for x in a triangle: x^2 + y^2 = -20x, x^2 + 20x = 0

A

alee

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x^2 + y^2 = -20x

Do you just isolate the x, by ignoring the y^2?

x^2 + 20x = 0

I'm lost
 
first off, where did these equations come from?

second, what is it you're trying to accomplish?
 
Triangle HAT is congruent to SHO
HT = x^2+y^2 and SO = -20x
y-x=20
find x
 
x^2 + y^2 = -20x

y - x = 20
y = x + 20 ... substitute (x+20) for y in the first equation ...

x^2 + (x + 20)^2 = -20x

solve for x, you'll get two solutions ... only one may be valid, then again maybe both solutions are valid. you'll have to check your solutions to see which makes sense in the context of the problem.
 
Alee, POST THE WHOLE ORIGINAL PROBLEM.

So far, makes no sense: x is negative; -10 or -20.
 
Thank you, I just totally missed the substitution, you know, can't see the forest for the trees!
 
x^2 + Y^2 = -20x y-x=20
x^2 +(20+x)^2 = -20x
x^2 + 400 + x^2 = -20x
2x^2 + 20x +400 = 0
2(x^2 +10x +200) = 0
then there is no solution because nothing mulitiplied together equals 200 while added together equals 10, is this correct?
 
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