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Let $a, b, c, d$ be real numbers between $-5$ and 5 such that $|a|=\sqrt{4-\sqrt{5-a}},|b|=\sqrt{4+\sqrt{5-b}}$,
$$
|c|=\sqrt{4-\sqrt{5+c}},|d|=\sqrt{4+\sqrt{5+d}}
$$
Then the product abcd is
Options:
$$
|c|=\sqrt{4-\sqrt{5+c}},|d|=\sqrt{4+\sqrt{5+d}}
$$
Then the product abcd is
Solution:
1254 Upvotes
Verified Answer
The correct answer is:
11
Given $|\mathrm{a}|=\sqrt{4-\sqrt{5-\mathrm{a}}}$
squaring
$\begin{array}{c}
a^{2}=4-\sqrt{5-a} \\
\Rightarrow a^{4}+16-8 a^{2}=5-a \\
\Rightarrow a^{4}-8 a^{2}+a+11=0
\end{array}$
Sirnilarly squaring other given equations
\& solving we can say that $a, b,-c,-d$ are roots of $x^{4}-8 x^{2}+x+11=0$
$\therefore$ product of roots
$a b(-c)(-d)=11$
abcd $=11$
squaring
$\begin{array}{c}
a^{2}=4-\sqrt{5-a} \\
\Rightarrow a^{4}+16-8 a^{2}=5-a \\
\Rightarrow a^{4}-8 a^{2}+a+11=0
\end{array}$
Sirnilarly squaring other given equations
\& solving we can say that $a, b,-c,-d$ are roots of $x^{4}-8 x^{2}+x+11=0$
$\therefore$ product of roots
$a b(-c)(-d)=11$
abcd $=11$
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