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The number of real roots of $|x|^2-5|x|+6=0$ is
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1704 Upvotes
Verified Answer
The correct answer is:
4
We have,
$$
|x|^2-5|x|+6=0
$$
Let
$$
|x|=y
$$
$$
\begin{aligned}
& \Rightarrow & y^2-5 y+6 & =0 \\
\Rightarrow & & (y-2)(y-3) & =0 \\
\Rightarrow & & y & =2,3 \\
\Rightarrow & & |x| & =2 \text { or }|x|=3 \\
\Rightarrow & & x & = \pm 2 \text { or } \pm 3
\end{aligned}
$$
$\therefore$ Number of real roots are 4.
$$
|x|^2-5|x|+6=0
$$
Let
$$
|x|=y
$$
$$
\begin{aligned}
& \Rightarrow & y^2-5 y+6 & =0 \\
\Rightarrow & & (y-2)(y-3) & =0 \\
\Rightarrow & & y & =2,3 \\
\Rightarrow & & |x| & =2 \text { or }|x|=3 \\
\Rightarrow & & x & = \pm 2 \text { or } \pm 3
\end{aligned}
$$
$\therefore$ Number of real roots are 4.
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