# find out the equation with the roots given

• Sep 26th 2009, 05:29 AM
Ilsa
find 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 AM
e^(i*pi)
Quote:

Originally Posted by Ilsa
i need to find out the equation, the roots are given:

-1+2√2 and -1-2√2
2 2

pls help

$[x - (\frac{-1+2\sqrt2}{2})] \times [x - (\frac{-1-2\sqrt2}{2})]$
• Sep 26th 2009, 07:27 AM
ramiee2010
you can also use this formula

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

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

Get my drift?
• Sep 28th 2009, 06:40 AM
Ilsa
i am using the formula bu can't get the equation, pls help by giving complete steps ~ thankyou.
• Sep 28th 2009, 06:47 AM
stapel
Quote:

Originally Posted by Ilsa
i am using the formula bu can't get the equation

Where are you stuck in the multiplication process?

Please be complete. Thank you! (Wink)
• Sep 28th 2009, 07:45 AM
ramiee2010
$x^2-(sum \ of\ roots)x+product \ of \ roots=0$
$\\ \alpha= \frac{-1+2\sqrt{2}}{2}\ \ \beta= \frac{-1-2\sqrt{2}}{2}$
$\alpha+ \beta= \frac{-1+2\sqrt{2}}{2}\ +\frac{-1-2\sqrt{2}}{2}\ = \frac{-1}{2}+\frac{2\sqrt{2}}{2}\ +\frac{-1}{2}-\frac{2\sqrt{2}}{2}$
$\alpha+ \beta=\frac{-2}{2} =\ -1$
$\\ \alpha \beta= \{ \frac{-1+2\sqrt{2}}{2}\ \} \{ \frac{-1-2\sqrt{2}}{2} \} \ = \{ \frac{1-2\sqrt{2}}{2}\ \} \{ \frac{1+2\sqrt{2}}{2} \}$
$\\ \alpha \beta=\frac {\{1^2-(2\sqrt{2})^2\}}{4} \ = \frac{(1-8)}{4}=\frac{-7}{4}$
therefore ,the equation is
$\\ x^2-(\alpha +\beta)x+\alpha. \beta=0$
$\\ x^2-(-1)x+\frac{-7}{4}=0$
$\\ 4x^2+4x-7=0$