# Which of the following has the maximum number of real roots?

##
#x^2-|x|-2=0#

#x^2-2|x|+3=0#

#x^2-3|x|+2=0#

#x^2+3|x|+2=0#

##### 1 Answer

#### Explanation:

Note that the roots of:

#ax^2+b abs(x)+c = 0#

are a subset of the union of the roots of the two equations:

#{ (ax^2+bx+c = 0), (ax^2-bx+c = 0) :}#

Note that if one of these two equations has a pair of real roots then so does the other, since they have the same discriminant:

#Delta = b^2-4ac = (-b)^2-4ac#

Further note that if

#x^2+3 abs(x)+2 >= 2#

so has no zeros.

Let's look at the other three equations in turn:

**1)**

#{ (0 = x^2-x-2 = (x-2)(x+1) \ => \ x in { -1, 2 }), (0 = x^2+x-2 = (x+2)(x-1) \ => \ x in { -2, 1 }) :}#

Trying each of these, we find solutions

**2)**

#Delta = b^2-4ac = (-2)^2-4(1)(3) = 4-12 = -8 < 0#

So this equation has no real roots.

**3)**

#{ (0 = x^2-3x+2 = (x-1)(x-2) \ => \ x in { 1, 2 }), (0 = x^2+3x+2 = (x+1)(x+2) \ => \ x in { -1, -2 }) :}#

Trying each of these, we find all are solutions of the original equation, i.e.

**Alternative method**

Note that real roots of

So to find which of the given equations has the most real roots is equivalent to finding which of the corresponding ordinary quadratic equations has most positive real roots.

A quadratic equation with two positive real roots has signs in the pattern

Of the given examples, only the second and third have coefficients in the pattern

We can discount the second equation

#0 = x^2-3x+2 = (x-1)(x-2)#

has two positive real roots, yielding