nonlinear inequalities in one variable-solution set in interval notation

Question

3x^2+16x<-5

Answer 1

Given inequality 3x^2+16x< -5

Add 5 to each side.

3x^2+16x+5 < -5+5

3x^2+16x+5 < 0

3x^2+15x+x+5 < 0

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

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

3x+1 < 0  and  x+5 < 0

3x < -1    and  x < -5

x < -1/3

The solution of inequality is x < -1/3 and x < -5.

Interval notation is {x/x<-1/3}

Comments

The inequality 3x ² + 16x < -5 solution is -5 < x  < -1/3.

Answer 2

Given inequality 3x^2+16x< -5

Add 5 to each side.

3x^2+16x+5 < -5+5

3x^2+16x+5 < 0

3x^2+15x+x+5 < 0

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

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

3x+1 < 0                         and  x+5 < 0

Subtract 1 from each side.    and  Subtract 5 from each side.

3x < -1                           and   x < -5

Divide to each side by 3.

3x/3 < -1/3

x < -1/3

The solution of inequality is x < -1/3 and x < -5.

Set notation is {x/x<-1/3}

The interval notation is negitive infinity to negitive 1/3.

 

Comments

The inequality 3x ² + 16x < -5 solution is -5 < x  < -1/3.

Answer 3

The inequality 3x ²+ 16x < - 5

3x ² + 16x + 5 < 0

3x ² + 16x  + 5 < 0

3x ²+ 15x + x + 5 < 0

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

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

Now, there are two ways this product could be less than zero.

One factor must be negative and one must be positive.

First situation:

3x + 1 < 0 and x + 5 > 0

3x  < -1 and x > -5

x  < - 1/3 and x  > -5

This tells us that -5 < x  < -1/3.

Second situation :

3x + 1 > 0 and x + 5 < 0

3x > -1 and x < -5

x > -1/3 and x < -5

There are NO values for which this situation is true.

Solution {x |x : -5 < x  < -1/3}

3x ² + 16x + 5 < 0 in the interval (-5, -1/3).

Comments

The solution set is {x |x : -5 < x  < -1/3}.

The interval notation form of solution set is (- 5, - 1/3), and closed interval represent the numbers - 5 and - 1/3 are not solutions of the inequality.