If $x^{2} + p x + q = 0$ and $x^{2} + q x + p = 0, (p \neq q)$ have a common root, show that $1 + p + q = 0$ ; show that their other roots are the roots of the equation $x^{2} + x + p q = 0$ .

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Solution

We have
Given function

$x^{2} + p x + q = 0$ ---- $(1)$

$x^{2} + q x + p = 0$ ---- $(2)$

By applying mathematical generality

$x = 1$ is one of the roots,
From which both equations, we get
$1 + p + q = 0$
from equation $1$
products of roots $= q$
Hence, roots of first equation $x = 1$ and $x = q$

Similarly, for equation $(2)$
$x = 1$ and $x = p$
the equation having roots $p$ and $q$ is
$x^{2} - (p + q) x + p q = 0$
On substituting the values
$x^{2} + x + p q = 0$

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