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
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;
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!