If the roots of the quadratic equation x2 + 2px + q = 0 are tan 30° and tan 15° respectively, then q is

tan⁡30∘+tan⁡15∘=−2ptan⁡30∘tan⁡15∘=q1=tan⁡45∘=tan⁡30∘+tan⁡15∘1−tan⁡30∘tan⁡15∘⇒ 1=−2p1−q⇒1−q=−2p⇒ q=1+2p

If the roots of the quadratic equation x2 + 2px + q = 0 are tan 30° and tan 15° respectively, then q is

tan⁡30∘+tan⁡15∘=−2ptan⁡30∘tan⁡15∘=q1=tan⁡45∘=tan⁡30∘+tan⁡15∘1−tan⁡30∘tan⁡15∘⇒ 1=−2p1−q⇒1−q=−2p⇒ q=1+2p

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If the roots of the quadratic equation x² + 2px + q = 0 are tan 30° and tan 15° respectively, then q is

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1 + 2p

1 – 2p

1 + p

1 – p

answer is D.

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Detailed Solution

$\tan 30^{\circ} + \tan 15^{\circ} = - 2 p$

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

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

$\Rightarrow 1 = \frac{- 2 p}{1 - q} \Rightarrow 1 - q = - 2 p$

$\Rightarrow q = 1 + 2 p$

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If the roots of the quadratic equation x2 + 2px + q = 0 are tan 30° and tan 15° respectively, then q is

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If the roots of the quadratic equation x² + 2px + q = 0 are tan 30° and tan 15° respectively, then q is