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Help please
If 2x = a + b + c show
(xa)^2 +(xb)^2 + (xc)^2+x^2 = a^2 +b^2+c^2.
 one year ago
 one year ago
Help please If 2x = a + b + c show (xa)^2 +(xb)^2 + (xc)^2+x^2 = a^2 +b^2+c^2.
 one year ago
 one year ago

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TeemoTheTerificBest ResponseYou've already chosen the best response.0
IF 2x=a+b+c Show that \[(xa)^{2}+(xb)^{2}+(xc)^{2} +x ^{2}=a ^{2}+b ^{2}+c ^{2}\]
 one year ago

TeemoTheTerificBest ResponseYou've already chosen the best response.0
it have no idea how to start!!!
 one year ago

darkb0nebeautyBest ResponseYou've already chosen the best response.0
You can use the Intermediate Value Theorem. Note that F is a polynomial and hence is differentiable everywhere (differentiable ==> continuous). I'll assume a < border doesn't really matter, but I do want a to be different from b. Consider the interval [a, b]. Note that f(a) = a, and f(b) = b. The number (a + b)/2 is between a and b (exactly half way in between!). The IVT says that if f takes on the values a and b it has to take on every value between them. It follows immediately that there exists c in (a, b) such that f(c) = (a + b)/2.
 one year ago

darkb0nebeautyBest ResponseYou've already chosen the best response.0
Does that help?
 one year ago

darkb0nebeautyBest ResponseYou've already chosen the best response.0
Well that is the only way i can explain it
 one year ago

TeemoTheTerificBest ResponseYou've already chosen the best response.0
whatever you just typed is very confusing as my question is clearly not a polynomial...
 one year ago

darkb0nebeautyBest ResponseYou've already chosen the best response.0
ok well sorry, bye..!
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
\[(xa)^2+(xb)^2+(xc)^2+x^2=x^22ax+a^2+x^22bx+b^2+x^22xc+c^2+x^2\\=4x^2+a^2+b^2+c^22ax2bx2cx\\=(2x)^2+a^2+b^2+c^22x(a+b+c)\\=(a+b+c)^2+a^2+b^2+c^2(a+b+c)(a+b+c)\\=(a+b+c)^2+a^2+b^2+c^2(a+b+c)^2\\=a^2+b^2+c^2 \]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
whoops, the \(+x^2\) wouldn't show.
 one year ago
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