Question #253d6

Oct 16, 2015

Here's what I got.

Explanation:

You know that two quadratic equations

$3 {x}^{2} + p x + 1 = 0 \text{ " " } \textcolor{p u r p \le}{\left(1\right)}$

and

$2 {x}^{2} + q x + 1 = 0 \text{ " " } \textcolor{p u r p \le}{\left(2\right)}$

have a common root and that the following relationship exists between $p$ and $q$

$2 {p}^{2} + 3 {q}^{2} - 5 p q + 1 = 0$

Now let's assume that $\textcolor{b l u e}{n}$ is the common root. Since that would imply that $\textcolor{b l u e}{n}$ is a root for both equations, you can say that

$3 \cdot {\textcolor{b l u e}{n}}^{2} + p \cdot \textcolor{b l u e}{n} + 1 = 0$

and

$2 \cdot {\textcolor{b l u e}{n}}^{2} + q \cdot \textcolor{b l u e}{n} + 1 = 0$

This is equivalent to saying that

$3 {n}^{2} + p n + \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} = 2 {n}^{2} + q n + \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}}$

Rearrange to get the value of $n$ in terms of $p$ and $q$

$3 {n}^{2} - 2 {n}^{2} + p n - q n = 0$

${n}^{2} + n \left(p - q\right) = 0$

$n \cdot \left(n + p - q\right) = 0 \implies \left\{\begin{matrix}n = 0 \\ n + p - q = 0 \iff n = q - p\end{matrix}\right.$

SInce $n = 0$ does not satify the original equations, you get that $n = q - p$.

Check to see if the given relationship is true, first for equation $\textcolor{p u r p \le}{\left(1\right)}$

$3 \cdot {\left(q - p\right)}^{2} + p \cdot \left(q - p\right) + 1 = 0$

$3 {q}^{2} - 6 p q + 3 {p}^{2} + p q - {p}^{2} + 1 = 0$

This is equal to

$2 {p}^{2} + 3 {q}^{2} - 5 p q + 1 = 0 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$

and the nfor equation $\textcolor{p u r p \le}{\left(2\right)}$

$2 {\left(q - p\right)}^{2} + q \left(q - p\right) + 1 = 0$

$2 {q}^{2} - 4 p q + 2 {p}^{2} + {q}^{2} - q p + 1 = 0$

This will once again give

$2 {p}^{2} + 3 {q}^{2} - 5 p q + 1 = 0 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$