Question #253d6

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Question #253d6

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Answer 1 · 2015-10-16T18:34:29

Here's what I got.

Explanation:

You know that two quadratic equations

$3 x^{2} + p x + 1 = 0 (1)$ $3x^2 + px + 1 = 0" " " "color(purple)((1))$

and

$2 x^{2} + q x + 1 = 0 (2)$ $2x^2 + qx + 1 = 0" " " "color(purple)((2))$

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

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

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

$3 \cdot n^{2} + p \cdot n + 1 = 0$ $3 * color(blue)(n)^2 + p * color(blue)(n) + 1 = 0$

and

$2 \cdot n^{2} + q \cdot n + 1 = 0$ $2 * color(blue)(n)^2 + q * color(blue)(n) + 1 = 0$

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

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

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Question #253d6

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