Question

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

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

Answer 1

Answer: (C)

$2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0$
$2|\sqrt{x}-3|+(\sqrt{x}-3+3)(\sqrt{x}-3-3)+6=0$
$2|\sqrt{x}-3|+(\sqrt{x}-3)^{2}-3=0$
$(\sqrt{x}-3)^{2}+2|\sqrt{x}-3|-3=0$
$(|\sqrt{x}-3|+3)(|\sqrt{x}-3|-1)=0$
$\Rightarrow |\sqrt{x}-3|=1,|\sqrt{x}-3|+3 \neq 0$
$\Rightarrow \sqrt{x}-3=\pm 1$
$\Rightarrow \sqrt{x}=4,2$
$x=16,4$