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Question-18818




Question Number 18818 by khamizan833@yahoo.com last updated on 30/Jul/17
Answered by daffa22 last updated on 30/Jul/17
(x^2 −6x+9)^(x^2 −4) =(x^2 −6x+9)^0   x^2 −4 = 0  (x−2)(x+2)=0  x = 2∨x = −2  for x∈R , x = { 2,−2 }
$$\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}\right)^{{x}^{\mathrm{2}} −\mathrm{4}} =\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}\right)^{\mathrm{0}} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}\:=\:\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$${x}\:=\:\mathrm{2}\vee{x}\:=\:−\mathrm{2} \\ $$$${for}\:{x}\in{R}\:,\:{x}\:=\:\left\{\:\mathrm{2},−\mathrm{2}\:\right\} \\ $$
Commented by khamizan833@yahoo.com last updated on 30/Jul/17
Good job brothee.  thanks for your solution.
$$\mathrm{Good}\:\mathrm{job}\:\mathrm{brothee}. \\ $$$$\mathrm{thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{solution}. \\ $$
Commented by daffa22 last updated on 30/Jul/17
sorry i made a mistake,  i don′t know what x∈R means,  thanks for your comment
$${sorry}\:{i}\:{made}\:{a}\:{mistake}, \\ $$$${i}\:{don}'{t}\:{know}\:{what}\:{x}\in{R}\:{means}, \\ $$$${thanks}\:{for}\:{your}\:{comment} \\ $$
Commented by FilupS last updated on 30/Jul/17
x∈R   means   x∈R  It means x is in the real set of numbers.  i.e.  Im(x)=0
$${x}\in{R}\:\:\:\mathrm{means}\:\:\:{x}\in\mathbb{R} \\ $$$$\mathrm{It}\:\mathrm{means}\:{x}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:{real}\:\mathrm{set}\:\mathrm{of}\:\mathrm{numbers}. \\ $$$$\mathrm{i}.\mathrm{e}.\:\:\mathrm{Im}\left({x}\right)=\mathrm{0} \\ $$
Answered by mrW1 last updated on 30/Jul/17
(1)  x^2 −4=0  ⇒x=±2  (2)  x^2 −6x+9=1  ⇒(x−3)^2 =1  ⇒x−3=±1  ⇒x=3±1  ⇒x=4,2  (3)  x^2 −6x+9=−1 and x^2 −4=even  ⇒no real solution    all solutions:  x=−2,2,4
$$\left(\mathrm{1}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\pm\mathrm{2} \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{9}=\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}−\mathrm{3}=\pm\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{3}\pm\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{4},\mathrm{2} \\ $$$$\left(\mathrm{3}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{9}=−\mathrm{1}\:\mathrm{and}\:\mathrm{x}^{\mathrm{2}} −\mathrm{4}=\mathrm{even} \\ $$$$\Rightarrow\mathrm{no}\:\mathrm{real}\:\mathrm{solution} \\ $$$$ \\ $$$$\mathrm{all}\:\mathrm{solutions}: \\ $$$$\mathrm{x}=−\mathrm{2},\mathrm{2},\mathrm{4} \\ $$
Commented by khamizan833@yahoo.com last updated on 30/Jul/17
agree with your solution.  thank mr.
$$\mathrm{agree}\:\mathrm{with}\:\mathrm{your}\:\mathrm{solution}. \\ $$$$\mathrm{thank}\:\mathrm{mr}. \\ $$

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