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# Help

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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
+102804
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I'd also like to see someone answer this.

May 26, 2019
#2
+28125
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Here is my attempt:

.

May 26, 2019
#3
+102804
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Thanks Alan,

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.

May 26, 2019
#4
+102434
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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

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