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

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

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

darkb0nebeauty
 one year ago
Best ResponseYou've already chosen the best response.0You 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.

darkb0nebeauty
 one year ago
Best ResponseYou've already chosen the best response.0Does that help?

darkb0nebeauty
 one year ago
Best ResponseYou've already chosen the best response.0Well that is the only way i can explain it

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

darkb0nebeauty
 one year ago
Best ResponseYou've already chosen the best response.0ok well sorry, bye..!

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.1\[(xa)^2+(xb)^2+(xc)^2+x^2=x^22ax+a^2+x^22bx+b^2+x^22xc+c^2+x^2\\=4x^2+a^2+b^2+c^22ax2bx2cx\\=(2x)^2+a^2+b^2+c^22x(a+b+c)\\=(a+b+c)^2+a^2+b^2+c^2(a+b+c)(a+b+c)\\=(a+b+c)^2+a^2+b^2+c^2(a+b+c)^2\\=a^2+b^2+c^2 \]

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.1whoops, the \(+x^2\) wouldn't show.
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