How do you find all zeros of the function #x² + 24 = –11x#?

2 Answers
Jun 4, 2016

Answer:

#x=-3color(white)("XXX")andcolor(white)("XXX")x=-8#

Explanation:

Re-writing the given equation as
#color(white)("XXX")x^2+11x+24=0#
and remembering that
#color(white)("XXX") (x+a)(x+b)=x^2+(a+b)x+ab#

We are looking for two values, #a# and #b# such that
#color(white)("XXX")a+b=11# and
#color(white)("XXX")ab=24#

with a bit of thought we come up with the pair #3# and #8#

So we can factor:
#color(white)("XXX")(x+3)(x+8)=0#

which implies either #x=-3# or #x=-8#

Jun 4, 2016

Answer:

x=-8 or x=-3

Explanation:

First you get the equivalent equation
#x^2+11x+24=0#
then you solve
#x=-11/2+-sqrt(11^2-4(24))/2#
#x=-11/2+-sqrt(25)/2#
#x=-11/2+-5/2#
so x=-8 or x=-3