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

1) 2
2) 3
3) 0
4) 1
in Mathematics by Bronze Status (9,486 points) | 15 views

1 Answer

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Best answer




$x^{2}+p x+q=0$

$\tan 30^{\circ}+\tan 15^{0}=-p$

$\tan 30^{\circ} \cdot \tan 15^{0}=q$

$\tan 45^{\circ}=\frac{\tan 30^{\circ}+\tan 15^{\circ}}{1-\tan 30^{0} \tan 15^{\circ}}=\frac{-p}{1-q}=1$


$\Rightarrow q-p=1 \quad \therefore 2+q-p=3$

by Bronze Status (9,486 points)

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