p and q are complex numbers such that |p|=2/5x+2, |q|=-3/4x-1, and |p+q|=2x+4. On what interval must x fall?

Answer 1

|p| + |q| > |p+q| per Cauchy inequality

(2/5)x + 2 + [(-3/4)x - 1] >= 2x+4

(2/5)x - (3/4)x + 2 - 1 >= 2x+4

(8-15)/20 x + 1 >= 2x+4

(-7/20) x + 1 >= 2x + 4

1 >= (2 - 7/20) x + 4

-3 >= (1 and 13/20) x

-3 >= (33/20)x

-60/33 >= x

-20/11 >= x

Answer 2

From the given information you can write four inequalities:

2/5x+2 >= 0,

-3/4x-1 >= 0,

2x+4 >= 0,

2x+4 <= (2/5x+2) + (-3/4x-1).

The first three are due to nonnegativity of the modulus, the fourth is due to the triangle inequality. All four must be satisfied. So by simplifying each and taking the intersection, we get the smallest possible interval containing x based on the given information.

The simplified inequalities are:

x >= -5,

x <= -4/3,

x >= -2,

x <= -60/47.

We may drop the first because it is implied by the third. We may also drop the fourth because it is implied by the second. Intersecting the remaining inequalities gives:

-2 <= x <= -4/3.