Find the ordered pair \((a,b)\) of real numbers such that the cubic polynomials \(x^3 + ax^2 + 11x + 6 = 0\) and \(x^3 + bx^2 + 14x + 8 = 0\) have two distinct roots in common.

Havingfun May 25, 2019

#3**0 **

Thanks Alan,

I have not finished looking at your answer but your start is similar to mine.

I did not bother with expanding

I used this

if:

\(ax^3+bx^2+cx+d=0\\ \text{and the roots are u,v and t}\\ then\\ u+v+t=-\frac{b}{a}\\ uv+ut+vt=+\frac{c}{a}\\ uvt=-\frac{d}{a}\)

The relevance is that I think **uvt=-8**

I got this far myself but then I got into a mess.

I am eager to examine the rest of your answer.

Melody May 26, 2019

#4**+1 **

x^3 + ax^2 + 11x + 6 = 0 (1)

x^3 + bx^2 + 14x + 8 = 0 (2)

By the Rational Roots Theorem, the possible zeroes for the first polynomial are ± [ 1, 2 , 3 , 6]

And for the second polynomial they are ± [ 1,2,3,4]

Subtract (1) from (2) and we get that

(b - a) x^2 + 3x + 2 = 0

Two possible shared roots of -1 and - 2 can be found if we let b - a = 1

So we have that

x^2 + 3x+ 2 = 0

(x + 1) ( x + 2) = 0

x = - 1 and x = -2

Now let x = -1 as a shared root

Then (-1)^3 + a(-1)^2 + 11(-1) + 6 = 0

-1 + a - 11 + 6 = 0

And a =6

Also

(-1)^3 + b(-1)^2 + 14(-1) +8 = 0

-1 + b -14 + 8 = 0

And b = 7

So testing (-2) as a root we have that

(-2)^3 + 6(-2)^2 + 11(-2) + 6 =

-8 + 24 - 22 + 6 =

0

And

(-2)^3 + 7(-2)^2 + 14(-2) + 8 =

-8 + 28 -28 + 8 =

0

So

(a,b ) = (6, 7)

Here is the graph to show that this is true :https://www.desmos.com/calculator/y69o2yts2n

CPhill May 26, 2019