How do you solve #(16-x^2)/(x^2-9)=0#?

Question

How do you solve #(16-x^2)/(x^2-9)>=0#?

Eva Adamson · Accepted Answer

Identify points and intervals at which this is satisfied to find solution:

#x in [-4, -3) uu (3, 4]#

Let #f(x) = (16-x^2)/(x^2-9) = ((4-x)(4+x))/((x-3)(x+3))# First note that if #x=+-4# then #16-x^2 = 0# and #x^2-9 != 0#, so these values of #x# are part of the solution set. Next note that if #x=+-3# then #x^2-9 = 0# and #16-x^2 != 0#, so #f(x)# is not even defined for these values of #x#, which are therefore not part of the solution set. Apart from these values of #x# the rational expression is non-zero and continuous, so is consistently positive or negative throughout each of the individual intervals: #(-oo, -4)#, #(-4, -3)#, #(-3, 3)#, #(3, 4)#, #(4, oo)# If #x < -4# or #x > 4# then #16-x^2 < 0# and #x^2-9 > 0#, so #f(x) < 0#. If #-4 < x < -3# or #3 < x < 4# then #16-x^2 > 0# and #x^2-9 > 0#, so #f(x) > 0#. If #-3 < x < 3# then #16-x^2 > 0# and #x^2-9 < 0#, so #f(x) < 0# So #f(x) >= 0# when #x in [-4, -3) uu (3, 4]#

Julia Adams · Answer

undefined First, find the critical points by setting the numerator and denominator equal to zero. Then, test the intervals created by the critical points to determine where the inequality is satisfied. The solutions are $x\leq -4, -3\leq x \leq 3, x\geq 4$.

How do you solve #(16-x^2)/(x^2-9)&gt;=0#?

How do you solve #(16-x^2)/(x^2-9)>=0#?