1. ## Algeb

$x^2+ax+1=0$ the quadratic equation have roots (a) and (b) and also they are not equals to other.a+b+c=0 prove this and thorough this prove that $x^2-ax+1$has the roots (b+a) (c+a)

2. ## Re: Algeb

Originally Posted by srirahulan
$x^2+ax+1=0$ the quadratic equation have roots (a) and (b) and also they are not equals to other.a+b+c=0 prove this and thorough this prove that $x^2-ax+1$has the roots (b+a) (c+a)
This question as stated is fundamentally flawed. That is unless a= $\frac{i\sqrt2}{2}$.
Do you mean to have complex coefficients?

yes

4. ## Re: Algeb

Originally Posted by srirahulan
yes
Well I gave you one of the roots, you find the other.

5. ## Re: Algeb

Originally Posted by srirahulan
$x^2+ax+1=0$ the quadratic equation have roots (a) and (b) and also they are not equals to other.a+b+c=0 prove this and thorough this prove that $x^2-ax+1$has the roots (b+a) (c+a)
a and b are the two roots of the equation but what is "c"?

Also you are using the same letter, a, to mean one of the roots and the coefficient of x. They cannot mean the same number:
if $(x- a)(x- b)= x^2- (a+b)x+ ab= x^2+ ax+ 1$ then a+ b= a, so b= 0, and ab= 1 which is impossible if b= 0.

6. ## Re: Algeb

Originally Posted by HallsofIvy
a and b are the two roots of the equation but what is "c"?
Also you are using the same letter, a, to mean one of the roots and the coefficient of x. They cannot mean the same number:
if $(x- a)(x- b)= x^2- (a+b)x+ ab= x^2+ ax+ 1$ then a+ b= a, so b= 0, and ab= 1 which is impossible if b= 0.
Yes indeed the a's are the same. Both $a~\&~b$ are complex numbers.