How do you solve $(x - 3) \frac{x - 4}{x - 5} {(x - 6)}^{2} < 0$ $(x-3)(x-4) / (x-5)(x-6)^2<0$ ?

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Answer 1 · 2018-02-09T20:44:23

$] - \infty, 3 [\cup] 4, 5 [$ $]-oo,3[uu]4,5[$

$(x - 3) \frac{x - 4}{x - 5} {(x - 6)}^{2}$ $(x−3)(x−4)/(x−5)(x−6)^2$ <0

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

Each term will be negative if xroot

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 $] - \infty, 3 [\cup] 4, 5 [$ $]-oo,3[uu]4,5[$

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