Question

The number of real solutions of the equation ${\left| x \right|^2} - 3\left| x \right| + 2 = 0$ is:
$\left( a \right)$ 4
$\left( b \right)$ 1
$\left( c \right)$ 3
$\left( d \right)$ 2

Answer 1

Hint: In this particular question use the concept that when x is less than zero then the modulus of x is open as negative and when x is greater than zero than the modulus of x is open as positive so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given equation:
${\left| x \right|^2} - 3\left| x \right| + 2 = 0$
Now equate modulus part to zero we have,
Therefore, x = 0
So when x < 0 modulus of x i.e. |x| open as negative and when x > 0 modulus of x i.e. |x| open as positive.
So when, x < 0, the given equation becomes
$ \Rightarrow {\left( { - x} \right)^2} - 3\left( { - x} \right) + 2 = 0$
$ \Rightarrow {x^2} + 3x + 2 = 0$
Now factorize the above equation we have,
$ \Rightarrow {x^2} + x + 2x + 2 = 0$
$ \Rightarrow x\left( {x + 1} \right) + 2\left( {x + 1} \right) = 0$
$ \Rightarrow \left( {x + 1} \right)\left( {x + 2} \right) = 0$
$ \Rightarrow x = - 1, - 2$
So when, x < 0 there are 2 real solutions.
Now when x > 0, the given equation becomes
$ \Rightarrow {\left( x \right)^2} - 3\left( x \right) + 2 = 0$
$ \Rightarrow {x^2} - 3x + 2 = 0$
Now factorize the above equation we have,
$ \Rightarrow {x^2} - x - 2x + 2 = 0$
$ \Rightarrow x\left( {x - 1} \right) - 2\left( {x - 1} \right) = 0$
$ \Rightarrow \left( {x - 1} \right)\left( {x - 2} \right) = 0$
$ \Rightarrow x = 1,2$
So when, x > 0 there are 2 real solutions.
So there total (2 + 2) = 4 real solutions.
So this is the required answer.

So, the correct answer is “Option A”.

Note: Whenever we face such type of questions equate the value which is inside the modulus to zero and solve for x, as above, we can also solve the quadratic equation (say $a{x^2} + bx + c = 0$) according to the quadratic formula which is given as $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, so on comparing first find out the values of a, b, and c and simply substitute the values in the above equation we will get the required solutions.