# equations

• Jun 25th 2010, 05:57 PM
Veronica1999
equations
For one root of ax(squared) + bx + c to be double the other, the coefficients a, b, c, must be related as follows:

a. 4b(squared) = 9c
b. 2b(squared) = 9ac
c. 2b(squared) = 9a
d. b(squared) - 8ac = 0
e. 9b(squared) = 2ac

I got the answer b by using (x-2)(x-4)=0.
I plugged in a=1 b= -6 c=8 but I think the problem wasn't meant to be solved that way. Is there a better way to do it?
• Jun 25th 2010, 06:42 PM
ramiee2010
For one root of $ax^2 + bx + c$ to be double the other

then

$\frac {-b- \sqrt{b^2-4ac} }{2a} = 2 \times \frac {-b+ \sqrt{b^2-4ac} }{2a}$
solving above equation we get
$2b^2=9ac$
• Jun 25th 2010, 07:08 PM
Also sprach Zarathustra
Another solution...
$ax^2+bx+c$

if $x_1$ and $x_2$ are two roots, we demand on $x_2$ to be $x_2=2x_1$

Now, sum of the two roots is $(*) x_1+x_2=3x_1=\frac{-b}{a}$ and their product is $x_1x_2=2x_1^2=\frac{c}{a}$

So, $x_1=\sqrt{\frac{c}{2a}}$ put this to $(*)$, and you get what you need to prove.