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TeemoTheTerific
 3 years ago
Help please
If 2x = a + b + c show
(xa)^2 +(xb)^2 + (xc)^2+x^2 = a^2 +b^2+c^2.
TeemoTheTerific
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0it have no idea how to start!!!

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

TeemoTheTerific
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok well sorry, bye..!

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