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TeemoTheTerific
Help please If 2x = a + b + c show (x-a)^2 +(x-b)^2 + (x-c)^2+x^2 = a^2 +b^2+c^2.
IF 2x=a+b+c Show that \[(x-a)^{2}+(x-b)^{2}+(x-c)^{2} +x ^{2}=a ^{2}+b ^{2}+c ^{2}\]
it have no idea how to start!!!
You can use the Intermediate Value Theorem. Note that F is a polynomial and hence is differentiable everywhere (differentiable ==> continuous). I'll assume a < b---order 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.
Well that is the only way i can explain it
whatever you just typed is very confusing as my question is clearly not a polynomial...
ok well sorry, bye..!
\[(x-a)^2+(x-b)^2+(x-c)^2+x^2=x^2-2ax+a^2+x^2-2bx+b^2+x^2-2xc+c^2+x^2\\=4x^2+a^2+b^2+c^2-2ax-2bx-2cx\\=(2x)^2+a^2+b^2+c^2-2x(a+b+c)\\=(a+b+c)^2+a^2+b^2+c^2-(a+b+c)(a+b+c)\\=(a+b+c)^2+a^2+b^2+c^2-(a+b+c)^2\\=a^2+b^2+c^2 \]
whoops, the \(+x^2\) wouldn't show.