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?

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- Sep 6th 2013, 06:26 AMTrefoil2727quadratic 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 AMtopsquarkRe: quadratic equation
- Sep 6th 2013, 08:45 AMred_dogRe: 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 $\displaystyle -\frac{3d}{2c}$