$\displaystyle x^2+ax+1=0$ 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 $\displaystyle x^2-ax+1$has the roots (b+a) (c+a)

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- May 11th 2012, 07:29 AMsrirahulanAlgeb
$\displaystyle x^2+ax+1=0$ 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 $\displaystyle x^2-ax+1$has the roots (b+a) (c+a)

- May 11th 2012, 07:52 AMPlatoRe: Algeb
- May 11th 2012, 09:23 PMsrirahulanRe: Algeb
yes

- May 12th 2012, 07:50 AMPlatoRe: Algeb
- May 15th 2012, 06:27 AMHallsofIvyRe: Algeb
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 $\displaystyle (x- a)(x- b)= x^2- (a+b)x+ ab= x^2+ ax+ 1$ then a+ b= a, so b= 0, and ab= 1 which is impossible if b= 0. - May 15th 2012, 07:01 AMPlatoRe: Algeb