Question

what is the value of x if
(x2-5x+6)/(x2+x+1)<0

Answer 1

Given the inequality,

$\longrightarrow\dfrac{x^2-5x+6}{x^2+x+1}<0$

The numerator $x^2-5x+6$ can be factorised as $(x-2)(x-3).$

$\longrightarrow\dfrac{(x-2)(x-3)}{x^2+x+1}<0\quad\quad\dots(1)$

In the case of denominator $x^2+x+1$ the discriminant is negative.

$\longrightarrow D=1^2-4\cdot1\cdot1$

$\longrightarrow D=-3$

Since $a=1>0$ and $D=-3<0,$

$\longrightarrow x^2+x+1>0$

The denominator is always positive, so numerator must be negative. Then (1) implies,

$\longrightarrow(x-2)(x-3)<0$

$\Longrightarrow\underline{\underline{x\in(2,\ 3)}}$

Hence $x$ can have the value of any real number exclusively between 2 and 3.

Answer 2

Answer:

It is the correct answer.

Step-by-step explanation:

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