Inequality question

Question

Inequality question
Find the complete set of inequation
2x+3 < √(-2-3x-x^2)

 

Please give the full answer 

Answer 1

You can only have real answers over the interval
where the radicand ... the part under the square root sign...
is ≥0
-x2-3x-2 = 0 would be a parabola opening downward.
Using the quadratic equation to find where this equals zero
will tell us where the parabola crosses the x-axis:
x = [-b ±√(b2-4ac)]/2a  with  a=-1, b=-3, c=-2
 
x = [3 ±√((-3)2-4(-1)(-2))]/2(-1)
x = (3±1)/(-2)
x = -2 or x=-1
Since this parabola touches the x axis in two places
and opens downward then the only area where the
radicand is greater than or equal to zero is for
-2≤x≤-1
All other values for x will result in a nonreal solution
 

Let's go back to the original inequality:

2x+3 < √(-2-3x-x^2)

Square both sides

4x² + 12x + 9 < -x²-3x-2

Add x²+3x+2 to both sides

5x² + 15x + 11 < 0

 

Again we are looking at a parabola.

This parabola opens upward since the

coefficient of the square term is positive.

We again need to use the quadratic

to find the points where this crosses

the x-axis.  All x values between those

two values will result in a negative answer.

 

5x² + 15x + 11 = 0

x = [-15 ±√(15²-4(5)(11))]/2(5)

x = (-15 ±√5)/10

 

This crosses the x-axis at the points

x = (-15+√5)/10 and x=(-15-√5)/10 

 

Working these two values out you get

x ≈ -1.3 and x≅ -1.7

 

Both of these values are within the

parameters set by the stricture on

the radicand.

 

Your inequality works for:

(-15-√5)/10 < x < (-15+√5)/10