Question

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?

• A. $11$
• B. $-11$
• C. $121$
• D. $-121$`

Answer 1

The correct option is C $121$

Let $a$ and $b$ be roots of an equation in $p$.

$\therefore |p|=\sqrt{4\pm \sqrt{5-p}}$

$\therefore p^2=4\pm \sqrt{5-p}$ (squaring both sides)

$\therefore p^4+16-8p^2=5-p$ (squaring both sides)

$\therefore p^4-8p^2+p+11=0$

Here, all 4 possibilities of roots is considered on opening the modulus of $a$ and $b$ and their product is 11.

Similarly, let $c$ and $d$ be roots of an equation in $q$.

$\therefore |q|=\sqrt{4\pm \sqrt{5+q}}$

$\therefore q^2=4\pm \sqrt{5+q}$ (squaring both sides)

$\therefore q^4+16-8q^2=5+q$ (squaring both sides)

$\therefore q^4-8q^2-q+11=0$

Similarly, product of all 4 possibilities of $q$ is $11.$

Hence, the answer is product of all possibilities of $p$ and $q$ i.e. $11\times 11=121$