The number of real solutions of the equation | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

If the set of all values of $a$, for which the equation $5 x ^3-15 x - a =0$ has three distinct real roots, is the interval $(\alpha, \beta)$, then $\beta-2 \alpha$ is equal to _____

The number of integral values of $m$ for which the quadratic expression, $(1 + 2m)x^2 -2(1+ 3m)x + 4(1 + m),$ $x\in R,$ is always positive, is

Let $p$ and $q$ be two real numbers such that $p+q=$ 3 and $p^{4}+q^{4}=369$. Then $\left(\frac{1}{p}+\frac{1}{q}\right)^{-2}$ is equal to

The number of distinct real roots of the equation $|\mathrm{x}+1||\mathrm{x}+3|-4|\mathrm{x}+2|+5=0$, is ...........

Suppose $a, b, c$ are positive integers such that $2^a+4^b+8^c=328$. Then, $\frac{a+2 b+3 c}{a b c}$ is equal to