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Problem: What will be the number of distinct real solution of $|x|^2-5|x|+6=0$ ?

Solution:

Case1: When $x<0$, then $x=-2,-3$

Case2: When $x>0$, then $x=2,3$

So number of solution is 4, am I right or making some mistake? As this a quadratic equation and the number of solutions must be equal to 2.

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3 Answers

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$|x|^2-5|x|+6=0 $ is aquadratic equation for $|x|$ ! (and not for $x$). We have

$|x|^2-5|x|+6=(|x|-3)(|x|-2) $, hence

$|x|^2-5|x|+6=0 $ $\iff $ $|x|=3$ or $|x|=2$ $\iff x \in \{-2,2,-3,3\}$

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This is a quadratic equation with absolute values..

You can factor this like this $$(|x|-3)(|x|-2)=0$$ So..either $$|x|=3 \Leftrightarrow x=\pm3$$ or$$|x|=2\Leftrightarrow x=\pm2$$ So your solutions are correct..

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You are correct.

It is a quadratic equation in $|x|$. Or you could set $y=|x|$, then: $$y^2-5y+6=0 \iff y=2 \vee y = 3$$ So with $y=|x|$, you get: $$|x|=2 \implies x = \pm 2 \quad\mbox{and}\quad |x|=3 \implies x = \pm 3$$

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