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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2023 · Mathematics · Quadratic Equation

Paragraph: Consider the following for the next two (02) items that...

Paragraph: Consider the following for the next two (02)(02) items that follow: A quadratic equation is given by (3+22)x2−(4+23)x+(8+43)=0(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?

Options

  1. A2(6−3+2−1)\sqrt{2}(\sqrt{6}-\sqrt{3}+\sqrt{2}-1)
  2. B2(6+3−2−1)\sqrt{2}(\sqrt{6}+\sqrt{3}-\sqrt{2}-1)
  3. C(6−3+2−1)(\sqrt{6}-\sqrt{3}+\sqrt{2}-1)
  4. D(6+3+2−1)(\sqrt{6}+\sqrt{3}+\sqrt{2}-1)

Correct answer

A. 2(6−3+2−1)\sqrt{2}(\sqrt{6}-\sqrt{3}+\sqrt{2}-1)

Step-by-step solution

GM=αβ=8+433+22=(8+43)(3−22)(3+22)(3−22)=4(6−42+33−26)=2(12−82+63−46)=2[(6−3)2+(2−1)2+2(6−3)(2−1)=2(6−3+2−1)2=2(6−3+2−1)\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).
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