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.
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.
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