Question

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

Answer 1

`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]}