# common roots of two cubic equations

• Jun 29th 2012, 12:50 AM
swordfish774
common roots of two cubic equations
(x^3)+(k*(x^2))+x+3=0;
(x^3)+(2*(x^2)+kx+2=0;

Find value of k, if the above two equations have a common root.
• Jun 29th 2012, 01:26 AM
biffboy
Re: common roots of two cubic equations
In both coefficient of x^3 is 1. 1st cubic ends in +3 and 2nd in +2. So any common factors can only be (x+1) or (x-1) giving roots -1 or 1
If x=1 is a common root 1+k+1+3=0 and 1+2+k+2=0 Both give k=-5
If x=-1 is a common root -1+k-1+3=0 and -1+2-k+2=0 These dont give the same k
So answer is k=-5 giving a common root of x=1
• Jun 29th 2012, 01:43 AM
emakarov
Re: common roots of two cubic equations
Quote:

Originally Posted by biffboy
In both coefficient of x^3 is 1. 1st cubic ends in +3 and 2nd in +2. So any common factors can only be (x+1) or (x-1) giving roots -1 or 1

Did you use the rational root theorem to conclude this? I believe there is another k that gives a common root, but it is irrational and hard to express...
• Jun 29th 2012, 01:53 AM
biffboy
Re: common roots of two cubic equations
Yes I did use that theorem.
• Jun 29th 2012, 01:57 AM
swordfish774
Re: common roots of two cubic equations
Quote:

Originally Posted by biffboy
In both coefficient of x^3 is 1. 1st cubic ends in +3 and 2nd in +2. So any common factors can only be (x+1) or (x-1) giving roots -1 or 1

Can you please explain this further?
• Jun 29th 2012, 01:58 AM
swordfish774
Re: common roots of two cubic equations
Quote:

Originally Posted by emakarov
Did you use the rational root theorem to conclude this? I believe there is another k that gives a common root, but it is irrational and hard to express...

What is the rational root theorem? Can you please link to an explanation with examples.
• Jun 29th 2012, 02:10 AM
emakarov
Re: common roots of two cubic equations
Quote:

Originally Posted by swordfish774
What is the rational root theorem? Can you please link to an explanation with examples.

See Wikipedia; it has an example. In this case, if p / q is a root of the first equation, then p | 3 and q | 1 (the notation "|" mean "divides"). Similarly, if p / q is a root of the second equation, then p | 2 and q | 1. Therefore, p and q are ±1 and the only possible rational roots are 1 and -1.

An alternative way is to subtract the second equation from the first one to get a quadratic equation and solve it for x. It is not as scary as it seems. This gives the solution shown above as well as another root as a function of k. Substituting this root into one of the equations gives a cubic equation on k, but its root is apparently irrational.