Expert: Scott A Wilson - 8/6/2009

Question
how to settle such question..
|2x-3|<|x+1|..the answer given is 2\3<x<4..
|3x=1|<(4x+3)2(square)

Answer
belle Country: Malaysia Category: Advanced Math Private: No Subject: modulus function Question: how to settle such question..
|2x-3|<|x+1|..the answer given is 2\3<x<4..
|3x=1|<(4x+3)2(square)

In this response, I will speak of x being in the interval (a,b).  That means a < x < b.


For the first equation, |2x-3|<|x+1|, the problem must be looked at in three regions.

The points in the middle of these three series are where 2x - 3 = 0 and x + 1 = 0.
Now x + 1 = 0 means that x = -1 and 2x - 3 = 0 means that x = 3/2 = 1.5.  

That means the series that must be looked at [1] (-∞, -1), [2] (-1, 3/2), and [3] (3/2, ∞).

In [1], x is in the interval (-∞, -1), both expressions are negatvive,
so the equations is -2x + 3 < -x - 1.
If we add 2x+1 to both sides we get 4 < x, but we know that x < 0, so this is not possible.

In [2], x is in the interval (-1, 3/2), we get -2x + 3 < x + 1.  
If 2x - 1 is added to both sides, the result is 2 < 3x, or x > 2/3.
Since for this to be true, x must be between -1 and 3/2 and we must have x > 2/3.
The interesction of these two intervals is (2/3, 3/2).

In [3], where x is in the interval (3/2, ∞).
Since both sides are positive, this breaks down to the equation 2x - 3 < x + 1.
Add 3 - x to both sides and get x < 4.
Since x has to be at least 3/2, the interval here is (3/2, 4).

Combining both of these intervals together gives us (2/3, 4).


For the second one, we have |3x - 1| < (4x+3)².

The only point to worry about is where 3x - 1 = 0, or 3x = 1, which is at x = 1/3.
This means we have to check to intervals.  
The first one is (-∞, 1/3) and the other one is (1/3,∞)

Lets first note that (4x+3)² = 16x² + 24x + 9.
{ squares are also done as ^2; 3^2 = 9, 5^2=25, 2^3 = 8, 2^5 = 32, etc. }

Next note that if x is 1/3 or greater, the line is 3x-1.  At 1/3, the line is 0 and (4x+3)² is (4/3 + 3)² = (13/9)² = 169/9, which is definitely higher.  The parabola increases ever faster, so there is no solution for x > 1/3.

If x < 1/3, the absolute value is 1 – 3x.  To find the difference between them, take
(4x+3)² - (1 – 3x) and get 16x² + 24x + 9 – 1 + 3x.  This is the same as 16x² + 27x + 8.

The two solutions to that equation can be found with the quadratic formula,
and both of them are less than 1/3.  The interval in which the function is negative is between these two values.

They are where x = (-b ± √(b²-4ac))/(2a), where a = 16, b = 27, and c = 8.
That’s is, (-27 ±√(27² - 4*16*8)/(2*16) = (-27 ± √(729 – 512))/32 = (-27 ± √217)/32.
Now the √217 is just less than 15, so when this is subtracted and added to –27 the answer is
–42/32 and –12/32.  These number are close to –5/4 and –3/8.  Note that –5/4 is a little less than –1 and
-3/8 is a little bigger than –1/2.

So the solution is all of the x values except for those between the two points that were just given.