find the set of values of x for which y is real if y=(((x−2)(x−1))/(x+2)) , x≠−2, x∈R


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Question Number 63534 by Rio Michael last updated on 05/Jul/19

$${find}\:{the}\:{set}\:{of}\:{values}\:{of}\:{x}\:{for}\:{which}\:{y}\:{is}\:{real}\:{if}\: \\ $$$$\:{y}=\frac{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)}{{x}+\mathrm{2}}\:,\:{x}\neq−\mathrm{2},\:{x}\in\mathbb{R} \\ $$
	Answered by MJS last updated on 05/Jul/19
	$the answer is already given: x∈R\{−2}$
	$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{already}\:\mathrm{given}:\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{2}\right\} \\ $$
	Commented by Rio Michael last updated on 05/Jul/19

	$${why}\:{is}\:{it}\:{so}? \\ $$
	Commented by Prithwish sen last updated on 05/Jul/19

	$$\mathrm{for}\:\mathrm{x}=−\mathrm{2}\:\mathrm{y}\:\mathrm{is}\:\mathrm{undefined} \\ $$
	Commented by MJS last updated on 05/Jul/19
	$∀x∈R: (x−2)(x−1)∈R ∀t∈R: (t/0) is not defined ∀t∈R: (t/(x+2))∈R ⇔ x+2≠0 ⇔ x≠−2 ∀x∈R\{−2}: (((x−2)(x−1))/(x+2))∈R$
	$$\forall{x}\in\mathbb{R}:\:\left({x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)\in\mathbb{R} \\ $$$$\forall{t}\in\mathbb{R}:\:\frac{{t}}{\mathrm{0}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined} \\ $$$$\forall{t}\in\mathbb{R}:\:\frac{{t}}{{x}+\mathrm{2}}\in\mathbb{R}\:\Leftrightarrow\:{x}+\mathrm{2}\neq\mathrm{0}\:\Leftrightarrow\:{x}\neq−\mathrm{2} \\ $$$$\forall{x}\in\mathbb{R}\backslash\left\{−\mathrm{2}\right\}:\:\frac{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)}{{x}+\mathrm{2}}\in\mathbb{R} \\ $$
	Commented by Rio Michael last updated on 11/Jul/19

	$${so}\:{the}\:{value}\:{of}\:{x}\:{for}\:{which}\:{y}\:{is}\:{real}\:{is}\:{the}\:{valueof}\:{x}\:{for}\:{which} \\ $$$${y}\:{is}\:{undefined}? \\ $$
	Commented by MJS last updated on 12/Jul/19
	$∀x∈R\{−2}: (((x−2)(x−1))/(x+2))∈R means ∀ for all x∈R\{−2} x real but x≠−2 : the following is true (((x−2)(x−1))/(x+2))∈R this expression is real$
	$$\forall{x}\in\mathbb{R}\backslash\left\{−\mathrm{2}\right\}:\:\frac{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)}{{x}+\mathrm{2}}\in\mathbb{R}\:\mathrm{means} \\ $$$$\forall\:\mathrm{for}\:\mathrm{all} \\ $$$${x}\in\mathbb{R}\backslash\left\{−\mathrm{2}\right\}\:{x}\:\mathrm{real}\:\mathrm{but}\:{x}\neq−\mathrm{2} \\ $$$$:\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{true} \\ $$$$\frac{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{1}\right)}{{x}+\mathrm{2}}\in\mathbb{R}\:\mathrm{this}\:\mathrm{expression}\:\mathrm{is}\:\mathrm{real} \\ $$
	Commented by Rio Michael last updated on 12/Jul/19

	$${i}\:{get}\:{it}\:{now}. \\ $$
	Commented by Rio Michael last updated on 12/Jul/19

	$${means}\:{if}\:{given}\:{something}\:{like} \\ $$$${y}\:=\:\left({x}+\mathrm{3}\right)\left({x}+\mathrm{6}\right)\:{to}\:{find}\:{the}\:{value}\:{of}\:{x}\:{for}\:{whichy}\:{isreal} \\ $$$$\:{x}\in\mathbb{R}? \\ $$
	Commented by MJS last updated on 12/Jul/19
	$yes y=(x−3)(x+5) x∈R ⇒ y∈R y=(1/((x−3)(x+5))) x∈R\{−5, 3} because y is not defined for these values y=((x−3)/(x+5)) x∈R\{−5}$
	$$\mathrm{yes} \\ $$$${y}=\left({x}−\mathrm{3}\right)\left({x}+\mathrm{5}\right) \\ $$$${x}\in\mathbb{R}\:\Rightarrow\:{y}\in\mathbb{R} \\ $$$${y}=\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{5}\right)} \\ $$$${x}\in\mathbb{R}\backslash\left\{−\mathrm{5},\:\mathrm{3}\right\}\:\mathrm{because}\:{y}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for} \\ $$$$\mathrm{these}\:\mathrm{values} \\ $$$${y}=\frac{{x}−\mathrm{3}}{{x}+\mathrm{5}} \\ $$$${x}\in\mathbb{R}\backslash\left\{−\mathrm{5}\right\} \\ $$
	Commented by Rio Michael last updated on 12/Jul/19

	$${thanks}\:{so}\:{much} \\ $$
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