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x-3-x-1-1-4-x-3-x-2-1-3-




Question Number 87637 by john santu last updated on 05/Apr/20
((∣x−3∣^(x+1) ))^(1/(4  ))  = ((∣x−3∣^(x−2) ))^(1/(3  ))
$$\sqrt[{\mathrm{4}\:\:}]{\mid\mathrm{x}−\mathrm{3}\mid^{\mathrm{x}+\mathrm{1}} }\:=\:\sqrt[{\mathrm{3}\:\:}]{\mid\mathrm{x}−\mathrm{3}\mid^{\mathrm{x}−\mathrm{2}} } \\ $$
Answered by TANMAY PANACEA. last updated on 05/Apr/20
∣x−3∣^((x+1)/4) =∣x−3∣^((x−2)/3)   ((x+1)/4)=((x−2)/3)  4x−8=3x+3  x=11
$$\mid{x}−\mathrm{3}\mid^{\frac{{x}+\mathrm{1}}{\mathrm{4}}} =\mid{x}−\mathrm{3}\mid^{\frac{{x}−\mathrm{2}}{\mathrm{3}}} \\ $$$$\frac{{x}+\mathrm{1}}{\mathrm{4}}=\frac{{x}−\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{4}{x}−\mathrm{8}=\mathrm{3}{x}+\mathrm{3} \\ $$$${x}=\mathrm{11} \\ $$
Commented by john santu last updated on 05/Apr/20
has three solution sir  x = 2; 4 ; 11
$$\mathrm{has}\:\mathrm{three}\:\mathrm{solution}\:\mathrm{sir} \\ $$$$\mathrm{x}\:=\:\mathrm{2};\:\mathrm{4}\:;\:\mathrm{11} \\ $$
Commented by TANMAY PANACEA. last updated on 05/Apr/20
ok
$${ok} \\ $$
Commented by Rasheed.Sindhi last updated on 05/Apr/20
What about x=3? Isn′t it root?
$${What}\:{about}\:{x}=\mathrm{3}?\:{Isn}'{t}\:{it}\:{root}? \\ $$
Commented by john santu last updated on 05/Apr/20
no
$$\mathrm{no} \\ $$
Commented by Rasheed.Sindhi last updated on 05/Apr/20
But x=3 satisfies the given eqn.  ^4 (√(∣x−3∣^(x+1) ))=^3 (√(∣x−3∣^(x−2) ))  ^4 (√(∣3−3∣^(3+1) ))=^3 (√(∣3−3∣^(3−2) ))  ^4 (√0^4 )=^3 (√0^1 )  ^4 (√0)=^3 (√0)      0=0
$${But}\:{x}=\mathrm{3}\:{satisfies}\:{the}\:{given}\:{eqn}. \\ $$$$\:^{\mathrm{4}} \sqrt{\mid{x}−\mathrm{3}\mid^{{x}+\mathrm{1}} }=\:^{\mathrm{3}} \sqrt{\mid{x}−\mathrm{3}\mid^{{x}−\mathrm{2}} } \\ $$$$\:^{\mathrm{4}} \sqrt{\mid\mathrm{3}−\mathrm{3}\mid^{\mathrm{3}+\mathrm{1}} }=\:^{\mathrm{3}} \sqrt{\mid\mathrm{3}−\mathrm{3}\mid^{\mathrm{3}−\mathrm{2}} } \\ $$$$\:^{\mathrm{4}} \sqrt{\mathrm{0}^{\mathrm{4}} }=\:^{\mathrm{3}} \sqrt{\mathrm{0}^{\mathrm{1}} } \\ $$$$\:^{\mathrm{4}} \sqrt{\mathrm{0}}=\:^{\mathrm{3}} \sqrt{\mathrm{0}} \\ $$$$\:\:\:\:\mathrm{0}=\mathrm{0} \\ $$

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