Solve the equation $x^3-11x^2+38x-40=0$, given that the ratio of two of its roots is $2:1$.
By hit and try, I can see that $x=2$ is a root.
Dividing the given cubic by $x-2$, I get a quadratic whose roots are $4,5$.
Can we solve this question without hit and try? Maybe by utilising the given ratio?
My Attempt:
Let $a,2a, b$ be the roots.
Thus, $2a^2b=40, 3a+b=11, 2a^2+3ab=38$.
Taking out $b$ from one equation and plugging in another isn't helping.
On
With $x_3=2x_2$, the equations $x_1+x_2+x_3=11$ and $x_1x_2+x_1x_3+x_2x_3=38$ gives $$7x_2^2-33x_2+38=0$$ Hence, $x_2=2$.
On
Since two of the zeroes of the polynomial are $ \ r \ $ and $ \ 2r \ $ (which tells us all three of the zeroes must be real, since any complex zeroes must be "conjugate"), we have $$ r^3 \ - \ 11r^2 \ + \ 38r \ - \ 40 \ \ = \ \ (2r)^3 \ - \ 11·(2r)^2 \ + \ 38·2r \ - \ 40 \ \ = \ \ 0 $$ $$ \Rightarrow \ \ 7r^3 \ - \ 33r^2 \ + \ 38r \ \ = \ \ r \ · \ (r \ - \ 2) \ · \ (7r \ - \ 19) \ \ = \ \ 0 \ \ . $$
Plainly, $ \ 0 \ $ cannot be one of the zeroes of the original cubic polynomial, and $ \ \frac{19}{7} \ $ is not a "candidate rational zero" of that polynomial. Hence $ \ 2 \ $ and $ \ 4 \ $ are two of the three real zeroes; synthetic or polynomial division shows the remaining zero to be $ \ 5 \ \ . $
This is equivalent to asking: if $ \ f(x) \ = \ x^3 - 11x^2 + 38x - 40 \ \ , \ $ what $ \ x-$intercept do $ \ f(x) \ $ and $ \ f(2x) \ $ have in common? (Those of $ \ f(2x) \ $ are $ \ 1 \ , \ 2 \ , \ $ and $ \ \frac52 \ \ . \ ) $
We have $$ f=(x-a)(x-2a)(x-b)=x^3 + x^2( - 3a - b) + xa(2a + 3b) - 2a^2b $$ Comparing coefficients with $x^3-11x^2+38x-40$ yields \begin{align*} 0 & = 3a+b-11,\\ 0 & = 2a^2+3ab-38,\\ 0 & = a^2b-20. \end{align*} It is not difficult to see that the solution is given by $(a,b)=(2,5)$. Indeed, let $b=11-3a$. Then the other two equations give $$ (7a - 19)(a - 2)=0,\; (3a^2 - 5a - 10)(a - 2)=0. $$ Since $7a=19$ contradicts the second equation, we need to have $a=2$.