(C) Given equation: $|x|^2-5|x|+6=0$Let $|x|=y$. Since $|x| \ge 0$,we have $y \ge 0$.The equation becomes $y^2-5y+6=0$.Factoring the quadratic: $(y-2)

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(C) Given equation: $|x|^2-5|x|+6=0$Let $|x|=y$. Since $|x| \ge 0$,we have $y \ge 0$.The equation becomes $y^2-5y+6=0$.Factoring the quadratic: $(y-2)(y-3)=0$.This gives $y=2$ or $y=3$.Substituting back $|x|=y$:Case $1$: $|x|=2 \Rightarrow x = \pm 2$.Case $2$: $|x|=3 \Rightarrow x = \pm 3$.Thus,the real roots are $x \in \{-3, -2, 2, 3\}$.Therefore,the number of real roots is $4$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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The number of real roots of $|x|^2-5|x|+6=0$ is

TS EAMCET 2016 All TS EAMCET

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

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