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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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