Answer:
[15-√226, 1] ∪ [29, 15+√226]
Explanation:
You want all solutions to |x^2 - 30x - 1| + |x^2 - 30x + 29| = 30.
Domains
The equation resolves to a piecewise equation whose domains are delimited by the points where the absolute value arguments are zero.
Left argument
The argument of the left absolute value function will be zero where ...
x^2 - 30x - 1 = 0
x = (-(-30) ±√((-30)² -4(1)(-1)))/(2(1) = 15 ± √226 ≈ {-0.0332964, 30.0332964}
Right argument
The argument of the right absolute value function will be zero where ...
x^2 -30x +29 = 0
(x -29)(x -1) = 0
x = {1, 29}
Piece-wise equations
-∞ < x < 15-√226
In this domain, both absolute value arguments are positive, so the equation is ...
x^2 -30x -1 +x^2 -30x +29 = 30
2x^2 -60x +28 = 30 . . . . . collect terms
x^2 -30x -1 = 0 . . . . . . . . . write in standard form
x = {15 -√226, 15 +√226} . . . . solutions to the equation
These solutions are not in the domain we have defined for this equation.
15-√226 ≤ x ≤ 1
In this domain, the left absolute value argument is negative, and the right argument is positive, so the equation is ...
-(x^2 -30x -1) +(x^2 -30x +29) = 30
30 = 30
All values of x in this domain (15-√226 ≤ x ≤ 1) are in the solution set.
1 < x < 29
In this domain, both absolute value arguments are negative, so the equation is ...
-(x^2 -30x -1) -(x^2 -30x +29) = 30
-2x^2 +60x -58 = 0 . . . . . . write in standard form
x^2 -30x +29 = 0 . . . . . . . divide by -2
The solutions to this are x={1, 29}, neither of which is in the domain we have defined for this equation.
29 ≤ x ≤ 15+√226
In this domain, the left absolute value argument is negative and the right argument is positive. The equation resolves to the same one as for the domain 15-√226 ≤ x ≤ 1: 30 = 30. Hence, all x-values in this domain (29 ≤ x ≤ 15+√226) are part of the solution set.
15+√226 < x
In this domain, both absolute value arguments are positive, so the equation is ...
x^2 -30x -1 = 0 . . . . . . . . . write in standard form
x = {15 -√226, 15 +√226} . . . . solutions to the equation
These solutions are not in the domain we have defined for this equation.
Solutions
In interval notation, the solution set for the equation is ...
[15-√226, 1] ∪ [29, 15+√226]