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(c) $a,\;b$ are roots of ${x^2} - 3x + p = 0$

$	herefore $$a + b = 3,\;ab = p$

$c,\;d$ are roots of ${x^2} - 12x + q = 0$

$	herefore $$c +

Vedclass · Accepted Answer

$17 : 15$

Vedclass · Answer

(c) $a,\;b$ are roots of ${x^2} - 3x + p = 0$

$	herefore $$a + b = 3,\;ab = p$

$c,\;d$ are roots of ${x^2} - 12x + q = 0$

$	herefore $$c + d = 12,\;cd = q$

$a,\;b,\;c,\;d$ are in GP.

$	herefore $$\frac{b}{a} = \frac{d}{c}$$ \Rightarrow $$\frac{{a + b}}{{a - b}} = \frac{{c + d}}{{c - d}}$

$ \Rightarrow $$\frac{{{{(a - b)}^2}}}{{{{(a + b)}^2}}} = \frac{{{{(c - d)}^2}}}{{{{(c + d)}^2}}}$$ \Rightarrow $$1 - \frac{{4ab}}{{{{(a + b)}^2}}} = 1 - \frac{{4cd}}{{{{(c + d)}^2}}}$

$ \Rightarrow $$\frac{{ab}}{{{{(a + b)}^2}}} = \frac{{cd}}{{{{(c + d)}^2}}}$$ \Rightarrow $$\frac{p}{9} = \frac{q}{{144}}$

$ \Rightarrow $$\frac{p}{1} = \frac{q}{{16}}$$ \Rightarrow $$\frac{p}{q} = \frac{1}{{16}}$$ \Rightarrow $$\frac{{p + q}}{{q - p}} = \frac{{17}}{{15}}$.

Trick : Let $a = 1,\;b = 2,\;c = 4,\;d = 8$, then
$p = 2,\;q = 32$ $ \Rightarrow $ $(q + p):(q - p) = 17:15$.

Let $a$ and $b$ be roots of ${x^2} - 3x + p = 0$ and let $c$ and $d$ be the roots of ${x^2} - 12x + q = 0$, where $a,\;b,\;c,\;d$ form an increasing G.P. Then the ratio of $(q + p):(q - p)$ is equal to

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