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
George C.
Jun 17, 2018

#x^2-3 abs(x)+2 = 0# with #4# real roots.

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 #a, b, c# all have the same sign then #ax^2+b abs(x) + c# will always take values of that sign when #x# is real. So in our examples, since #a=1#, we can immediately note that:

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

so has no zeros.

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

1) #x^2-abs(x)-2 = 0#

#{ (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 #x in { -2, 2 }#

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

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

So this equation has no real roots.

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

#{ (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. #x in { -2, -1, 1, 2 }#

Alternative method

Note that real roots of #ax^2+b abs(x)+c = 0# (where #c != 0#) are positive real roots of #ax^2+bx+c = 0#.

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 #+ - +# or #- + -#. In our example the first sign is always positive.

Of the given examples, only the second and third have coefficients in the pattern #+ - +#.

We can discount the second equation #x^2-2 abs(x) + 3 = 0# since its discriminant is negative, but for the third equation we find:

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

has two positive real roots, yielding #4# roots of the equation #x^2-3 abs(x)+2 = 0#

Related questions
Impact of this question
1840 views around the world
You can reuse this answer
Creative Commons License