Inequation being solved by table of signs

Case of a product

Example 1 : that is to say the inequation

x^3+6x^2+12x \ Ge -8 \,

To solve this type of inequations per table of signs, one gathers all in the first member to have zero in the second then one factorizes the first member obtained.

This thanks to the rule:

to know the sign of a product, it is enough to seek that of each one of its factors, then to deduce that from it from the product thanks to the rule of the signs .

Here, there is

x^3+6x^2+12x+8 \ Ge 0 \,

then

(x+2) ^3 \ Ge 0 \,

according to the remarkable identity

(a+b)^3=a^3+3a^2b+3ab^2+b^3 \,

To solve this inequation amounts seeking the sign of $(x+2) ^3 \,$ , i.e. that of $x+2 \,$ .

There is then the table of signs according to:

|--- |sign $x+2 \,$ | |--- |sign $(x+2) ^3 \,$ | |}

One concludes from it that the whole of the solutions of this inequation is:

Case of a quotient

Example 2 : That is to say the inequation

\ frac {1-2x} {x-3} \ Ge 0

The rule seen higher for a product is valid also for a quotient, in condition of having checked for which (S) value (S) this quotient does not exist. Here, it is not necessary that $x-3=0 \,$ thus one does not need only $x=3 \,$ .

Then one makes the table of signs according to: |--- |sign $1-2x \,$ | |--- |sign $x-3 \,$ | |--- |sign $\ frac {1-2x} {x-3} \,$ | |} The whole of the solutions is thus: $\ left$

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