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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.

 May 25, 2019
 #1
avatar+106885 
0

I'd also like to see someone answer this.

 May 26, 2019
 #2
avatar+28349 
+3

Here is my attempt:

 

.

 May 26, 2019
 #3
avatar+106885 
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.   laugh

 May 26, 2019
 #4
avatar+106514 
+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

 

 

cool cool cool

 May 26, 2019
edited by CPhill  May 26, 2019
edited by CPhill  May 26, 2019

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