The value of #x# satisfying the equation #abs(abs(abs(x^2-x+4)-2)-3)=x^2+x-12# is?

1 Answer
Jul 30, 2018

Answer:

#x=11/2#

Explanation:

Given:

#abs(abs(abs(x^2-x+4)-2)-3) = x^2+x-12#

Note that:

#x^2+x-12 = (x+4)(x-3)#

which is non-negative (as required) when #x <= -4# or #x >= 3#

Also note that:

#x^2-x+4 = (x-1/2)^2+15/4 > 2" "# for all #x in RR#

So:

#abs(abs(x^2-x+4)-2) = abs((x-1/2)^2+15/4-2) = (x-1/2)^2+7/4#

If #x <= -4# or #x >= 3# then #(x-1/2)^2 >= (5/2)^2 = 25/4 > 3#

So when #x <= -4# or #x >= 3#:

#abs(abs(abs(x^2-x+4)-2)-3) = x^2-x-1#

and we want to solve:

#x^2-x-1 = x^2+x-12#

Subtracting #x^2-x-12# from both sides, this becomes:

#11 = 2x#

Hence:

#x = 11/2 > 3#

graph{(y-abs((x^2-x+2)-3))(y-(x^2+x-12)) = 0 [-7.7, 12.3, -6, 32]}