First slide

Theory of equations

Question

If the roots of the quadratic equation $x^{2} + p x + y = 0$ are are tan 30° and tan 15° respectively, then the value of $2 + q - p$ is,

Moderate

A

2
B

3
C

0
D

1

Solution

It is given that tan 30° and tan 15° are roots of the
given equation.
$∴ \tan 30^{\circ} + \tan 15^{\circ} = - p$ and $\tan 30^{\circ} \tan 15^{\circ} = q$
Now, $\tan 45^{\circ} = \frac{\tan 30^{\circ} + \tan 15^{\circ}}{1 - \tan 30^{\circ} \tan 15^{\circ}}$
$\Rightarrow 1 = \frac{- p}{1 - q} \Rightarrow 1 - q = - p \Rightarrow q - p = 1 \Rightarrow q - p + 2 = 3$

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