Question

Let S = { $xϵR:x\ge 0$ and $2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0$ } . Then S

• A. contains exactly two elements
• B. contains exactly four elements
• C. Is an empty set
• D. contains exactly one elements

Answer 1

The correct option is A contains exactly two elements

Given,

$S=\{x\in R:x\ge 0,2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0\}$

we have,

$x\ge 0$ and

$2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0$

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$\Rightarrow 2(-(\sqrt{x}-3))+\sqrt{x}(\sqrt{x}-6)+6=0$

factorising the above equation, we get,

$(6-\sqrt{x})(2-\sqrt{x})=0$

$6-\sqrt{x}=0\Rightarrow x=36$

$2-\sqrt{x}=0\Rightarrow x=4$

$\therefore x=4,36$

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$\Rightarrow 2(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}-6)+6=0$

upon expansion of equation, we get,

$x-4\sqrt{x}=0$

$\Rightarrow \sqrt{x}=4$

$\therefore x=0,16$

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Upon merging the overlapping intervals, we get,

$x=4,16$ exactly $2$ elements