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

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1
Feb 9, 2018

]-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[

Then teach the underlying concepts
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#### Explanation

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#### Explanation:

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dagatho Share
Feb 9, 2018

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

#### Explanation:

The inequality is asking when the equation is negative. ${\left(x - 6\right)}^{2}$ is always positive, so it doesn't affect the sign of the equation.

$\frac{x - 4}{x - 5}$ 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, $\frac{x - 4}{x - 5}$ is positive, and when $\frac{x - 4}{x - 5}$ 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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