Recent content by larry91

Yes! The smallest value for a! and there is ≥. Sorry! Using Rolle's theorem is quite simple to solve it... I don't want to solve it 'cause I need a solution for it... I just want to find an approach through we can solve it using algebraic methods!

Hi everybody! I have to solve the following equation using algebraic methods: a=? such as 2x + ax > 3x + 4x for any x from ℝ. I solved it using analytical methods. Using The Rolle Theorem, but I don't want to solve it in that way! The solution for a is 6

Ok! Thanks! :smile:

I tried this: We can assume that F(x1,x2,x3) is a symetric ecuation; => F(x1,x2,x3) = F(xs(1),xs(2),xs(3)), where s is a permutation of the {1, 2, 3} set; F(x1,x2,x3) = F(x2,x1,x3) => \frac{{x_2 }}{{x_1 }} + \frac{{x_3 }}{{x_2 }} + \frac{{x_1 }}{{x_3 }} = \frac{{x_3 }}{{x_1 }} + \frac{{x_1...

Ok! I try this: \frac{x_2}{x_1}+\frac{x_3}{x_2}+\frac{x_1}{x_3} = \frac{x_2^2x_3}{x_1x_2x_3}+\frac{x_3^2x_1}{x_1x_2x_3}+\frac{x_1^2x_2}{x_1x_2x_3}= x22x3+x32x1 + x12x2; And from here I tried a lot of different approaches but I didn't succeeded to solve it!

So, I'll give you an example! f(x) = x3+x-1; To calculate: \frac{1}{x_1+x_2}+\frac{1}{x_1+x_3}+\frac{1}{x_2+x_3} From Viete's relations => x1+x2+x3 = 0 => => x1+x2 = -x3 x1x2+x1x3+x2x3=1 x1x2x3=1 So, we have to calculate : -\frac{1}{x_3}-\frac{1}{x_2}-\frac{1}{x_1} =...

Sorry once again for mistake! f(x) = x3+x-1; where x1, x2, x3 are the solutions of the equation f(x) = x3+x-1; To calculate : \frac{x_2}{x_1}+\frac{x_3}{x_2}+\frac{x_1}{x_3}. f(x) is a polynomial function;

Yes! Sorry! f(x1) = f(x2) = f(x3) = 0;

From Viete's relations => x_1+x_2+x_3=0,x_1x_2+x_1x_3+x_2x_3=1,x_1x_2x_3=1; From these relations you should to calculate S; x_1, x_2, x_3 are the solution of the F ecuation;

Either f(x)=x3+x-1; To calculate: \frac{x_2}{x_1}+\frac{x_3}{x_2}+\frac{x_1}{x_3}. I tried a lot of approaches but I didn't get the solution! May you help me please?

Ok! I try this, but I dislike this method: Either 'P mod Q' the rest of the P at Q rapport; (x+2)2 mod Q = -(x+3) (x+2)3 mod Q = (x+2)2(x+2) mod Q = (-(x+3)(x+2)) mod Q = 1 => the debris of (x+2)m at Q are repeated from 3 to 3; (x+3)2 mod Q = x+2 (x+3)3 mod Q = ((x+2)(x+3)) mod Q = -1...

Hi everybody! I have this problem: Either P = (X+2)m+(X+3)n and Q = x2+5x+7; Determine m, n such that Q | P;( m, n = ? (Q divide P)); May you help me please? Thank You!