i need to find out the equation, the roots are given:

-1+2√2and-1-2√2

2 2

pls help

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- Sep 26th 2009, 05:29 AMIlsafind out the equation with the roots given
i need to find out the equation, the roots are given:

__-1+2√2__and__-1-2√2__

2 2

pls help - Sep 26th 2009, 05:42 AMe^(i*pi)
- Sep 26th 2009, 07:27 AMramiee2010
you can also use this formula

x^2 - (sum of roots) x + (product of roots ) = 0 - Sep 26th 2009, 07:38 AMWilmer
x^2 + x + C = 0

x = [-1 +- sqrt(1 - 4C)] / 2

Get my drift? - Sep 28th 2009, 06:40 AMIlsa
i am using the formula bu can't get the equation, pls help by giving complete steps ~ thankyou.

- Sep 28th 2009, 06:47 AMstapel
- Sep 28th 2009, 07:45 AMramiee2010
$\displaystyle x^2-(sum \ of\ roots)x+product \ of \ roots=0$

$\displaystyle \\ \alpha= \frac{-1+2\sqrt{2}}{2}\ \ \beta= \frac{-1-2\sqrt{2}}{2}$

$\displaystyle \alpha+ \beta= \frac{-1+2\sqrt{2}}{2}\ +\frac{-1-2\sqrt{2}}{2}\ = \frac{-1}{2}+\frac{2\sqrt{2}}{2}\ +\frac{-1}{2}-\frac{2\sqrt{2}}{2}$

$\displaystyle \alpha+ \beta=\frac{-2}{2} =\ -1$

$\displaystyle \\ \alpha \beta= \{ \frac{-1+2\sqrt{2}}{2}\ \} \{ \frac{-1-2\sqrt{2}}{2} \} \ = \{ \frac{1-2\sqrt{2}}{2}\ \} \{ \frac{1+2\sqrt{2}}{2} \}$

$\displaystyle \\ \alpha \beta=\frac {\{1^2-(2\sqrt{2})^2\}}{4} \ = \frac{(1-8)}{4}=\frac{-7}{4}$

therefore ,the equation is

$\displaystyle \\ x^2-(\alpha +\beta)x+\alpha. \beta=0$

$\displaystyle \\ x^2-(-1)x+\frac{-7}{4}=0$

$\displaystyle \\ 4x^2+4x-7=0$