Welcome :: Homework Help and Answers :: Mathskey.com

Recent Visits

    
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,435 questions

17,804 answers

1,438 comments

778,743 users

Solve:

0 votes

55/(x+16) > -x?

and graph the inequality

asked Oct 4, 2014 in ALGEBRA 1 by anonymous

1 Answer

0 votes

The rational inequality is 55/(x + 16) > - x.

write the inequality in general form with the rational expression on the left and zero on the right.

[55/(x + 16)] + x > 0

[55 + x(x + 16)]/(x + 16) > 0

(55 + x2 + 16x)/(x + 16) > 0

(x2 + 11x + 5x + 55)/(x + 16) > 0

[(x + 11)(x + 5)]/(x + 16) > 0

The zeros (the x-values for which its numerator is zero) and its undefined values (the x-values for which its denominator is zero) of the rational expression are called key numbers.

Numerator is zero, x = - 11 and x = - 5.

Denominator is zero, x = - 16.

The key numbers are x = - 16, x = - 11 and x = - 5. So, the polynomial’s test intervals are (-∞, -16), (- 16, - 11), (- 11, - 5) and (- 5, ∞).

In each test interval, choose a representative x-value and evaluate the polynomial.

Test Interval   x-value   Polynomial Value [{(x + 11)(x + 5)}/(x + 16)]  Conclusion

(-∞, -16)         x = -17    [{(-17 + 11)(-17 + 5)}/(-17 + 16)] = - 72 < 0             Negative

(- 16, - 11)      x = -15    [{(-15 + 11)(-15 + 5)}/(-15 + 16)] = 40 > 0               Positive

(- 11, - 5)        x = -10    [{(-10 + 11)(-10 + 5)}/(-10 + 16)] = - 5/6 < 0            Negative

(- 5, ∞)           x = 0       [{(0 + 11)(0 + 5)}/(0 + 16)] = 55/16 > 0                    Positive

From this we can conclude that the inequality is satisfied on the open intervals (- 16, - 11) and (- 5, ∞). So, the solution set is (- 16, - 11) U (- 5, ∞) and its graph is

answered Oct 4, 2014 by casacop Expert

Related questions

asked Oct 17, 2018 in ALGEBRA 1 by anonymous
asked Sep 5, 2014 in ALGEBRA 1 by anonymous
asked Jun 25, 2014 in ALGEBRA 1 by anonymous
asked May 11, 2014 in ALGEBRA 1 by anonymous
...