Let a, b, c, d be real numbers between -5 and | vedclass

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(A) Given equations are:$|a|=\sqrt{4-\sqrt{5-a}}$$|b|=\sqrt{4+\sqrt{5-b}}$$|c|=\sqrt{4-\sqrt{5+c}}$$|d|=\sqrt{4+\sqrt{5+d}}$Squaring the first equatio

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$11$

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(A) Given equations are:$|a|=\sqrt{4-\sqrt{5-a}}$$|b|=\sqrt{4+\sqrt{5-b}}$$|c|=\sqrt{4-\sqrt{5+c}}$$|d|=\sqrt{4+\sqrt{5+d}}$Squaring the first equation: $a^2 = 4 - \sqrt{5-a} \implies a^2 - 4 = -\sqrt{5-a}$.Squaring again: $(a^2 - 4)^2 = 5 - a \implies a^4 - 8a^2 + 16 = 5 - a \implies a^4 - 8a^2 + a + 11 = 0$.Similarly,for $b, c, d$,we observe that $a, b, -c, -d$ are the roots of the polynomial equation $x^4 - 8x^2 + x + 11 = 0$.By Vieta's formulas,the product of the roots of $x^4 + 0x^3 - 8x^2 + x + 11 = 0$ is given by the constant term,which is $11$.Thus,$a \cdot b \cdot (-c) \cdot (-d) = 11 \implies abcd = 11$.

Let $a, b, c, d$ be real numbers between $-5$ and $5$ such that $|a|=\sqrt{4-\sqrt{5-a}}$,$|b|=\sqrt{4+\sqrt{5-b}}$,$|c|=\sqrt{4-\sqrt{5+c}}$,and $|d|=\sqrt{4+\sqrt{5+d}}$. Then,the product $abcd$ is

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