Re: Prove that A²+3B²=4.
anemone said:
Problem:
$$p(x)=4x^3-2x^2-15x+9$$
$$q(x)=12x^3+6x^2-7x+1$$
Let A be the largest root of $$p(x)$$ and B the largest root of $$q(x)$$. Show that $$A^2+3B^2=4$$
Hi all, this problem seems to be a little harder than usual for me and I hope someone will give me hint on how to solve it...
Thanks in advance. :)
You're right, this looks like an unusually hard problem. I have a sort of solution, but I don't believe it can be the best one.
If $\alpha,\ \beta,\ \gamma$ are the roots of $x^3 - px^2 + qx - r$, then the polynomial whose roots are $\alpha^2,\ \beta^2,\ \gamma^2$ is $x^3 - (p^2-2q)x^2 + (q^2-2pr)x - r^2$. The polynomial whose roots are $3\alpha,\ 3\beta,\ 3\gamma$ is $x^3 - 3px^2 + 9qx - 27r$; and the polynomial whose roots are $4-\alpha,\ 4-\beta,\ 4-\gamma$ is $(4-x)^3 -p(4-x)^2 + q(4-x) - r.$ Using those techniques, you can check that if $A,\ A',\ A''$ are the roots of $4x^3-2x^2-15x+9$, then the polynomial with roots $4-A^2, 4-A'^2, 4-A''^2$ is $x^3 - \frac{17}4x^2 + \frac{37}{16}x -\frac3{16}$. If $B,\ B',\ B''$ are the roots of $12x^3 + 6x^2 - 7x + 1$, then the polynomial with roots $3B^2, 3B'^2, 3B''^2$ is exactly the same: $x^3 - \frac{17}4x^2 + \frac{37}{16}x -\frac3{16}$.
This shows that $4-A^2$ must be one of $3B^2, 3B'^2, 3B''^2$, but it doesn't tell you which of those three it is. You know for example that $B$ is the largest root of the polynomial $q(x)$, but that does not mean that $B^2$ is the largest root of the polynomial with roots $3B^2, 3B'^2, 3B''^2$. In fact, each of the given polynomials $p(x)$, $q(x)$ has a negative root that is larger than either of its positive roots. So when you square the roots, the negative root has the largest square. This means that the
middle root of $x^3 - \frac{17}4x^2 + \frac{37}{16}x -\frac3{16}$ is the one that is equal to $4-A^2$ and $3B^2$, from which it follows that $A^2+3B^2=4.$
Afterthought: bad notation! After writing this out, I realized that I should not have used the letters $p,\ q,\ r$ for the coefficients in a cubic polynomial, because the given polynomials $p(x)$ and $q(x)$ already used the letters $p$ and $q$.
