The number of real solutions of the equation | vedclass

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Question

(d) Given $|x{|^2} - 3|x| + 2 = 0$

Here we consider two cases $viz.\,\,x < 0$and $x > 0$

Case I : $x < 0$ This gives ${x^2} + 3x + 2 =

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(d) Given $|x{|^2} - 3|x| + 2 = 0$ Here we consider two cases $viz.\,\,x < 0$and $x > 0$ Case I : $x < 0$ This gives ${x^2} + 3x + 2 = 0$ ==> $(x + 2)(x + 1) = 0\,\, \Rightarrow x = - 2, - 1$ Also $x = - 1, - 2$satisfy $x < 0,$so $x = - 1$, $-2$ is solution in this case. Case II : $x > 0$. This gives ${x^2} - 3x + 2 = 0$ ==> $(x - 2)(x - 1) = 0\,\, \Rightarrow x = 2,1$, so $x = 2$,$ 1$ is solution in this case. Hence the number of solutions are four i.e. $x = - 1,\,1,\,2,\, - 2$ Aliter : $|x{|^2} - 3|x| + 2 = 0$ ==> $(|x| - 1)(|x| - 2) = 0$ ==> $|x| = 1$and $|x| = 2$==> $x = \pm 1,x = \pm 2$.

The number of real solutions of the equation $|x{|^2}$-$3|x| + 2 = 0$ are

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