# quadratic equation

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• Sep 6th 2013, 06:26 AM
Trefoil2727
quadratic equation
Given that the equation x^2 +cx+d=0 and x^2 +3cx+4d=0 have a common root, express d in term of c

a+b=-c, a+q=-3c
ab=d aq=4d
aq-ab=3d
a(q-b)=3d...(1)
q-b=-2c...(2)
subst, a=-3d/2c

a^2 +ca+d=0
(-3d/2c)^2 +c(-3d/2c)+d=0
d=(2/9)(c^2)

am I right?
• Sep 6th 2013, 07:25 AM
topsquark
Re: quadratic equation
Quote:

Originally Posted by Trefoil2727
Given that the equation x^2 +cx+d=0 and x^2 +3cx+4d=0 have a common root, express d in term of c

a+b=-c, a+q=-3c
ab=d aq=4d
aq-ab=3d
a(q-b)=3d...(1)
q-b=-2c...(2)
subst, a=-3d/2c

a^2 +ca+d=0
(-3d/2c)^2 +c(-3d/2c)+d=0
d=(2/9)(c^2)

am I right?

Put the value for d into your two quadratic equations and factor. They share a common solution x = -3c.

-Dan
• Sep 6th 2013, 08:45 AM
red_dog
Re: quadratic equation
Substract de first ecuation from the second:
$\displaystyle 2cx+3d=0$
The common root of the two quadratics is also the root of the last equation.
So, the common root is $\displaystyle x=-\frac{3d}{2c}$
Now replace x from any of the two quadratics with $\displaystyle -\frac{3d}{2c}$