How do you solve #(x-3)(x-4) / (x-5)(x-6)^2<0#?

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Feb 9, 2018

Answer:

#]-oo,3[uu]4,5[#

Explanation:

#(x−3)(x−4)/(x−5)(x−6)^2#<0

The polynomial has 4 roots:3,4,5 and 6.

Each term will be negative if x<root
equal to zero if x=root
Bigger than zero if x>root

A table of + and - would be almost impossible to do here, so let's make a list:

if x<3: all terms are negative so the multiplication gives negative

if x=3 polynomial equal to 0

if x in ]3,4[, (x-3) is positive and (x-4) (x-5) are negative, so the multiplication will give positive

if x=4 polynomial equal to 0

if x in ]4,5[, only (x-5) is negative so the multiplication will give negative

if x=5 polynomial is undefined

if x in ]5,+oo[, all terms are positive so multiplication will give positive

(x−6)^2 will always positive except when x=6.

The result is #]-oo,3[uu]4,5[#

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Write your answer here...
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dagatho Share
Feb 9, 2018

Answer:

The answer is #x<3# or (#x>4# and #x<5#)

Explanation:

The inequality is asking when the equation is negative. #(x-6)^2# is always positive, so it doesn't affect the sign of the equation.

#(x-4)/(x-5)# is negative when #x>4# and #x<5#, because that is when only one side of the fraction is negative.

#x-3# is negative when #x<3#.

When #x-3# is negative, #(x-4)/(x-5)# is positive, and when #(x-4)/(x-5)# is negative, #x-3# is positive. So, the equation is negative when #x<3# or when #x>4# and #x<5#.

Therefore, the answer is #x<3# or #x>4# and #x<5#.

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