Ive got a problem and my maths test is 2mw! gaaaaaa
The roots of the equation X^2 + (P-10)X = 10P are a and a + 4 . Find the possible values of P
a is actually the aplha sign thingy but i dunno how to type it
Hello,
You have the equation : X^2 + (P-10)X = 10P
You know that $\displaystyle \alpha$ (click on it to see how i wrote it) and $\displaystyle \alpha+4$ are solutions.
This means that :
$\displaystyle \alpha^2+(P-10) \alpha=10P$
$\displaystyle (\alpha+4)^2+(P-10)(\alpha+4)=10P$
I dont get how you derive to this:
$\displaystyle \alpha^2+(P-10) \alpha=10P$
$\displaystyle (\alpha+4)^2+(P-10)(\alpha+4)=10P$
Why is X replaced by $\displaystyle \alpha$
Sorry, but im really dumb at roots. have a phobia of it somehow... Never manage to understand it ><
And thanks for showing me the way to do e alpha thing! ^_^
Well, if you have a trinomial :
ax˛+bx+c=0, $\displaystyle \alpha$ and $\displaystyle \beta$ two roots of the equation, you have :
$\displaystyle \alpha+\beta=\frac{-b}{a}$
and
$\displaystyle \alpha \beta=\frac{c}{a}$
Here, a=1, b=P-10, c=-10P, $\displaystyle \beta=\alpha+4$
And just replace ^^
The system to solve will be :
$\displaystyle \alpha + \alpha +4=\frac{-(P-10)}{1}=10-P$
$\displaystyle \alpha (\alpha+4)=\frac{-10P}{1}=-10P$
The first equation will give : $\displaystyle 2 \alpha = 6-P$
Then replace it in the second equation to get P
Sorry, i didn't understand the title as i should have