What is the number of solutions of the equation #abs(x^2-2)=absx#?

2 Answers
Dec 27, 2016

Answer:

#abs(x^2-2)=abs(x)# has #color(green)(4)# solutions

Explanation:

#abs(x^2-x)=abs(x)#
#rArr#
#color(white)("XXX"){:("Either",," or ",), (,x^2-2=x,,x^2-2=-x), (,x^2+x-2=0,,x^2+x-2=0), (,(x+2)(x-1)=0,,(x-2)(x+1)=0), (,x=-2 or +1,,x=+2 or -1) :}#

So there are 4 possible solutions:
#color(white)("XXX")x in {-2, -1, +1, +2}#

Dec 27, 2016

Answer:

Graph reveals solutions # x = +-1 and x = +-2#..

Explanation:

The graphs #y = |x| and y =|x^2-2|# intersect at #x = +-1 and x = +-2#.

So, these are the solutions of #(x-2|=|x|#.

Of course, algebraically, these solutions can be obtained, using

piecewise definitions, sans #|...|# symbol.

Note of caution: In general, graphical solutions are approximations

only.

graph{(y-|x|)(y-|x^2-2|)=0x^2 [-5, 5, -2.5, 2.5]}