(A) Given equation is $x^2 - 5|x| + 6 = 0$.Since $|x|^2 = x^2$,we can write the equation as $|x|^2 - 5|x| + 6 = 0$.Let $|x| = t$,where $t \ge 0$. Then

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(A) Given equation is $x^2 - 5|x| + 6 = 0$.Since $|x|^2 = x^2$,we can write the equation as $|x|^2 - 5|x| + 6 = 0$.Let $|x| = t$,where $t \ge 0$. Then the equation becomes $t^2 - 5t + 6 = 0$.Factoring the quadratic: $(t - 2)(t - 3) = 0$.This gives $t = 2$ or $t = 3$.Case $1$: $|x| = 2 \implies x = 2, -2$.Case $2$: $|x| = 3 \implies x = 3, -3$.Thus,the solutions are $x \in \{2, -2, 3, -3\}$.There are $4$ solutions in total.

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The number of solutions for the equation $x^2 - 5|x| + 6 = 0$ is

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A
$4$
B
$3$
C
$2$
D
$1$

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Type of microbes	Scientific name	Commercial product
Bacterium	$a$	Clot buster
$b$	Aspergillus niger	Citric acid
Fungus	Trichoderma polysporum	$c$
Bacterium	$d$	Butyric acid

The number of solutions for the equation $x^2 - 5|x| + 6 = 0$ is

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Identify the blank spaces $a, b, c$ and $d$ given in the following table and select the correct answer.

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Bacterium $d$ Butyric acid