# Given the equation: x^2 - |x| = mx (x+1), m - real number, if M = {m | the equation has got exactly 3 real and distinct roots}. Then M=?. There are five answers to choose from: (-inf, -1] ; (-1,...

Given the equation: x^2 - |x| = mx (x+1), m - real number, if M = {m | the equation has got exactly 3 real and distinct roots}. Then M=?.

There are five answers to choose from: (-inf, -1] ; (-1, 1) ; (2, +inf) ; the empty set; the set of real numbers.

At first, I was tempted to say that a second degree equation can only have two roots, but I thought about it more thoroughly and realised that there is that |x| that may change the situation, because there are two possibilities: x positive, or negative. Does it make any sense?

I don't know what to say, I find this exercise quite challenging and I would appreciate it very much if you could help me!

Have a nice day!

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Rearrange to isolate |x| :

|x| = x^2 - mx(x+1)

Now we have two equations:

(1) x = x^2 - mx(x+1) x>0

(2) -x = x^2 - mx(x+1) x<0

Solving (1)

(1-m)x^2 - mx - x = 0

(1-m)x^2 - (1+m)x = 0

(1-m)x - (1+m) = 0

Therefore:

x = (1+m)/(1-m) when x > 0; therefore:

(1+m)/(1-m) > 0

1 +m > 0

m > -1

Solving (2)

(1-m)x^2 - mx + x = 0

(1-m)x^2 - (m-1)x = 0

(1-m)x - (m-1) = 0

Therefore:

x = (m-1)/(1-m) when x < 0; therefore:

(m-1)/(1-m) < 0

m-1 < 0

m < 1

Therefore the solution is (-1,1)

First let us assume that x is negative, the equation x^2-|x|=mx(x+1) is equal to

x^2-x=m(-x)(-x+1)

=>x^2-x=mx^2-mx

=>x^2(m-1)-x(m-1)=0

=>x^2-x=0 or x(x-1)=0 therefore the roots are x=0 and x=1

As we had assumed that x is negative we see that x=1 cannot be a root.

Next, assume x is positive

x^2-|x|=mx(x+1)

=>x^2-x=m(x)(x+1)

=>x^2-x=mx^2+mx

=>x^2(m-1)+x(m+1)=0

=>x{x(m-1)+m+1)=0

The roots here are x=0 and x=-(m+1)/(m-1)

Therefore if we consider the general case where x can be either positive or negative, the equation x^2-|x|=mx(x+1) cannot have 3 real and distinct roots. Therefore the correct option is the empty set.