Question

Solve the following question

${x}^{2} - 5 |x| + 6 = 0$
and hence find number of solutions of this equation.
​

Answer 1

Two case
If x>o and x
Case 1)x>0
Then |x|=x
So solve equation x^2-5x+6=0
We get x=2,x=3

Case2)x<0
Then |x |=(-x)
Solve eqxn. x^2-5(-x)+6=0
x^2+5x+6=0
X= -2 ,X= -3


Got it

Answer 2

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\: {x}^{2} - 5 |x| + 6 = 0$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {x}^{2} = { |x| }^{2} \: }}}$

So, using this, we get

$\rm :\longmapsto\: { |x| }^{2} - 5 |x| + 6 = 0$

Now, its a quadratic in |x|, so using Concept of Splitting of middle terms, we get

$\rm :\longmapsto\: { |x| }^{2} - 3 |x| - 2 |x| + 6 = 0$

$\rm :\longmapsto\: |x|( |x| - 3) - 2( |x| - 3) = 0$

$\rm :\longmapsto\:( |x| - 3)( |x| - 2) = 0$

$\rm\implies \: |x| = 2 \: \: or \: \: |x| = 3$

$\rm\implies \: x \: = \: \pm \: 2 \: \: or \: \: x \: = \: \pm \: 3$

So,

$\bf\implies \:x = - 2, \: - 3, \: 2, \: 3$

$\bf\implies \:Number \: of \: solutions \: = \: 4$

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More to Know :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac