Solve the following function

Question

1. |2x+3|=1
2. |2x+3|<|x-2|

Answer 1

Hi Pius,

 

1) |2x+3| = 1

2x+ 3 = 1 or -(2x+3) = 1

2x+ 3 = 1 or -2x - 3 = 1

2x = -2    or 2x = -4

x = -1   or   x = -2

x = {-2, -1}

 

2) |2x+3|<|x-2|

We can take the square root of both sides since both sides have an absolute value. Thus, we get rid of the absolute value symbol because the square of any number is always non-negative just like absolute value.

(2x + 3)^2 < (x - 2)^2

4x^2 + 12x + 9 < x^2 - 4x + 4

3x^2 + 16x + 5 < 0

(3x + 1)(x + 5) < 0

x = -1/3 or x = -5, the expression equals zero. We need to find the interval when it is negative. Let's test.

when x < -1/3, the expression is negative.

when x > -1/3, the expression is positive.

when x < -5, the expression is positive.

when x > -5, the expression is negative.

So, the interval that |2x+3|<|x-2| is satisfied is -5 < x < -1/3.