(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.

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- Jun 29th 2012, 01:50 AMswordfish774common 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, 02:26 AMbiffboyRe: 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, 02:43 AMemakarovRe: common roots of two cubic equations
- Jun 29th 2012, 02:53 AMbiffboyRe: common roots of two cubic equations
Yes I did use that theorem.

- Jun 29th 2012, 02:57 AMswordfish774Re: common roots of two cubic equations
- Jun 29th 2012, 02:58 AMswordfish774Re: common roots of two cubic equations
- Jun 29th 2012, 03:10 AMemakarovRe: common roots of two cubic equations
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.