Q200 / 270 · NDA 2023 (Phase 1) · Mathematics

Paragraph: Consider the following for the next two $(02)$ items that follow: A quadratic equation is given by $(3+2 \sqrt{2}) x^2-(4+2 \sqrt{3}) x+(8+4 \sqrt{3})=0$
Question: What is the GM of the root of the equation?

  • Correct answer
    A
    $\sqrt{2}(\sqrt{6}-\sqrt{3}+\sqrt{2}-1)$
  • B
    $\sqrt{2}(\sqrt{6}+\sqrt{3}-\sqrt{2}-1)$
  • C
    $(\sqrt{6}-\sqrt{3}+\sqrt{2}-1)$
  • D
    $(\sqrt{6}+\sqrt{3}+\sqrt{2}-1)$

Solution

$\mathrm{GM}=\sqrt{\alpha \beta}=\sqrt{\frac{8+4 \sqrt{3}}{3+2 \sqrt{2}}}$ $=\sqrt{\frac{(8+4 \sqrt{3})(3-2 \sqrt{2})}{(3+2 \sqrt{2})(3-2 \sqrt{2})}}$ $=\sqrt{4(6-4 \sqrt{2}+3 \sqrt{3}-2 \sqrt{6})}$ $=\sqrt{2(12-8 \sqrt{2}+6 \sqrt{3}-4 \sqrt{6})}$ $=\sqrt{2\left[(\sqrt{6}-\sqrt{3})^2+(\sqrt{2}-1)^2+2(\sqrt{6}-\sqrt{3})(\sqrt{2}-1)\right.}$ $=\sqrt{2(\sqrt{6}-\sqrt{3}+\sqrt{2}-1)^2}=\sqrt{2}(\sqrt{6}-\sqrt{3}+\sqrt{2}-1)$.

Hence, the correct answer is option: 'A'.

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