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BITSAT 2024Paper 1
Exam
BITSAT
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Quadratic Equation

BITSAT 2024 · Paper 1 · Mathematics · Quadratic Equation

If 15^ ^ and 30^ ^ are the roots of the equation x^2+p x+q=0 , then...

If tan⁡15∘\tan 15^{{}^{\circ}} and tan⁡30∘\tan 30^{{}^{\circ}} are the roots of the equation x2+px+q=0x^2+p x+q=0, then pq=p q=

Options

  1. A63+103\frac{6 \sqrt{3}+10}{\sqrt{3}}
  2. B10−633\frac{10-6 \sqrt{3}}{3}
  3. C10+633\frac{10+6 \sqrt{3}}{3}
  4. D10−633\frac{10-6 \sqrt{3}}{\sqrt{3}}

Correct answer

B. 10−633\frac{10-6 \sqrt{3}}{3}

Step-by-step solution

∵tan⁡15∘=tan⁡(45−30)∘=3−13+1∴tan⁡15∘+tan⁡30∘=−p⇒p=−43(3+1)q=tan⁡15∘×tan⁡30∘=3−13(3+1)⇒q=3−13(3+1)∴p⋅q=−43(3+1)×(3−1)3(3+1)⇒p⋅q=−4(3−1)3(3+1)2=10−633⇒p⋅q=10−633⇒\because \quad \tan 15^{{}^{\circ}}=\tan (45-30)^{{}^{\circ}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\therefore \quad \tan 15^{{}^{\circ}}+\tan 30^{{}^{\circ}}=-p\Rightarrow \quad p=\frac{-4}{\sqrt{3}(\sqrt{3}+1)}q=\tan 15^{{}^{\circ}} \times \tan 30^{{}^{\circ}}=\frac{\sqrt{3}-1}{\sqrt{3}(\sqrt{3}+1)} \Rightarrow q=\frac{\sqrt{3}-1}{\sqrt{3}(\sqrt{3}+1)}\therefore p \cdot q=\frac{-4}{\sqrt{3}(\sqrt{3}+1)} \times \frac{(\sqrt{3}-1)}{\sqrt{3}(\sqrt{3}+1)}\Rightarrow \quad p \cdot q=\frac{-4(\sqrt{3}-1)}{3(\sqrt{3}+1)^2}=\frac{10-6 \sqrt{3}}{3}\Rightarrow p \cdot q=\frac{10-6 \sqrt{3}}{3}\Rightarrow Option (2) is correct.
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