Question #253d6

You know that two quadratic equations

`3x^2 + px + 1 = 0" " " "color(purple)((1))`

and

`2x^2 + qx + 1 = 0" " " "color(purple)((2))`

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

`2p^2 + 3q^2 -5pq + 1 = 0`

Now let's assume that `color(blue)(n)` is the common root. Since that would imply that `color(blue)(n)` is a root for both equations, you can say that

`3 * color(blue)(n)^2 + p * color(blue)(n) + 1 = 0`

and

`2 * color(blue)(n)^2 + q * color(blue)(n) + 1 = 0`

This is equivalent to saying that

`3n^2 + pn + color(red)(cancel(color(black)(1))) = 2n^2 + qn + color(red)(cancel(color(black)(1)))`

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

`3n^2 - 2n^2 + pn - qn = 0`

`n^2 + n(p-q) = 0`

`n * (n + p-q) = 0 implies {(n=0), (n + p - q = 0 <=> n = q - p) :}`

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 `color(purple)((1))`

`3 * (q-p)^2 + p * (q-p) + 1 = 0`

`3q^2 - 6pq + 3p^2 + pq - p^2 + 1 = 0`

This is equal to

`2p^2 + 3q^2 - 5pq + 1 = 0color(white)(x)color(green)(sqrt())`

and the nfor equation `color(purple)((2))`

`2(q-p)^2 + q(q-p) + 1 = 0`

`2q^2 - 4pq + 2p^2 + q^2 - qp + 1 = 0`

This will once again give

`2p^2 + 3q^2 - 5pq + 1 = 0color(white)(x)color(green)(sqrt())`