Q1:jipmat 2025QA › IndicesHardQA · MCQ

$\sqrt{2+\sqrt{3}} \times \sqrt{2+\sqrt{2+\sqrt{3}}} \times \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$ is equal to

• A1
• B2
• C4
• D$\sqrt{6}$

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The Setup: This looks like a terrifying infinite radical, but it's actually a satisfying chain reaction. We work from right to left, using the difference of squares identity: $(a+b)(a-b)=a^2-b^2$. Step 1: Multiply the last two terms together. They are identical except for the sign in the middle. Let $X=\sqrt{2+\sqrt{3}}$. The last two terms are $\sqrt{2+\sqrt{X}}$ and $\sqrt{2-\sqrt{X}}$. $$\sqrt{2+\sqrt{X}} \times \sqrt{2-\sqrt{X}}=\sqrt{(2)^2-(\sqrt{X})^2}=\sqrt{4-X}$$ Substitute $X$ back in: $$\sqrt{4-(2+\sqrt{3})}=\sqrt{2-\sqrt{3}}$$ Step 2: Now our expression is shorter. Multiply this new result by the second term from the original equation. Let's drop the visual noise. $$\sqrt{2+\sqrt{3}} \times (\sqrt{2+\sqrt{2+\sqrt{3}}} \times \sqrt{2-\sqrt{3}})$$ Apply the difference of squares again! Let $Y=\sqrt{3}$. $$\sqrt{2+\sqrt{Y}} \times \sqrt{2-\sqrt{Y}}=\sqrt{4-Y}=\sqrt{4-(2+\sqrt{3})}=\sqrt{2-\sqrt{3}}$$ Step 3: Final stage. Multiply this with the very first term. $$\sqrt{2+\sqrt{3}} \times \sqrt{2-\sqrt{3}}$$ $$\sqrt{(2)^2-(\sqrt{3})^2}=\sqrt{4-3}=\sqrt{1}=1$$ Final Answer: 1