The value of xx satisfying the equation abs(abs(abs(x^2-x+4)-2)-3)=x^2+x-12∣∣∣∣∣∣x2−x+4∣∣−2∣∣−3∣∣=x2+x−12 is?
1 Answer
Jul 30, 2018
Explanation:
Given:
abs(abs(abs(x^2-x+4)-2)-3) = x^2+x-12∣∣∣∣∣∣x2−x+4∣∣−2∣∣−3∣∣=x2+x−12
Note that:
x^2+x-12 = (x+4)(x-3)x2+x−12=(x+4)(x−3)
which is non-negative (as required) when
Also note that:
x^2-x+4 = (x-1/2)^2+15/4 > 2" "x2−x+4=(x−12)2+154>2 for allx 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
So when
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
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]}