The value of xx satisfying the equation abs(abs(abs(x^2-x+4)-2)-3)=x^2+x-12x2x+423=x2+x12 is?

1 Answer
Jul 30, 2018

x=11/2x=112

Explanation:

Given:

abs(abs(abs(x^2-x+4)-2)-3) = x^2+x-12x2x+423=x2+x12

Note that:

x^2+x-12 = (x+4)(x-3)x2+x12=(x+4)(x3)

which is non-negative (as required) when x <= -4x4 or x >= 3x3

Also note that:

x^2-x+4 = (x-1/2)^2+15/4 > 2" "x2x+4=(x12)2+154>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]}