Thread: 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.

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

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...

Yes I did use that theorem.

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?

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.

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.