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Question-145666

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Question Number 145666 by mathdanisur last updated on 07/Jul/21
Answered by Rasheed.Sindhi last updated on 07/Jul/21
∣x+1∣+(√(x^2 −4x−4))+∣x−z∣+∣z−3∣=4 ..... ....
$$\mid{x}+\mathrm{1}\mid+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{4}}+\mid{x}−{z}\mid+\mid{z}−\mathrm{3}\mid=\mathrm{4} \\ $$$$….. \\ $$$$…. \\ $$
Commented by mathdanisur last updated on 07/Jul/21
yes Ser, please
$${yes}\:{Ser},\:{please} \\ $$
Commented by mathdanisur last updated on 07/Jul/21
Ser, answer: {(2;z) and 2≤z≤3}
$${Ser},\:{answer}:\:\left\{\left(\mathrm{2};{z}\right)\:{and}\:\:\mathrm{2}\leqslant{z}\leqslant\mathrm{3}\right\} \\ $$
Commented by Rasheed.Sindhi last updated on 07/Jul/21
mathdanisur sir/madam your solution: (2,z) & 2≤z≤3 doesn′t satisfy the given equation. (√(x^2 +2x+1))+(√(x^2 −4x−4))+(√(x^2 −2xz+z^2 ))+(√(z^2 −6z+9))=4 For example: x=2,z=2: (√(2^2 +2(2)+1))+(√(2^2 −4(2)−4))+(√(2^2 −2(2)(2)+2^2 )) +(√(2^2 −6(2)+9))=^(?) 4 3+2i(√2)+0+1≠4 If the solution is correct then pl check the question: May be it is as: (√(x^2 +2x+1))+(√(x^2 −4x+4))+(√(x^2 −2xz+z^2 ))+(√(z^2 −6z+9))=4? In this case the solution will be (2,z) & z=2 or 3 i-e (2,2) or (2,3)
$${mathdanisur}\:{sir}/{madam} \\ $$$$\:{your}\:{solution}: \\ $$$$\left(\mathrm{2},{z}\right)\:\&\:\mathrm{2}\leqslant{z}\leqslant\mathrm{3}\: \\ $$$${doesn}'{t}\:{satisfy}\:{the}\:{given}\:{equation}. \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{4}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{xz}+{z}^{\mathrm{2}} }+\sqrt{{z}^{\mathrm{2}} −\mathrm{6}{z}+\mathrm{9}}=\mathrm{4} \\ $$$${For}\:{example}: \\ $$$${x}=\mathrm{2},{z}=\mathrm{2}: \\ $$$$\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2}\right)+\mathrm{1}}+\sqrt{\mathrm{2}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2}\right)−\mathrm{4}}+\sqrt{\mathrm{2}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}\right)\left(\mathrm{2}\right)+\mathrm{2}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\sqrt{\mathrm{2}^{\mathrm{2}} −\mathrm{6}\left(\mathrm{2}\right)+\mathrm{9}}\overset{?} {=}\mathrm{4} \\ $$$$\mathrm{3}+\mathrm{2}{i}\sqrt{\mathrm{2}}+\mathrm{0}+\mathrm{1}\neq\mathrm{4} \\ $$$${If}\:{the}\:{solution}\:{is}\:{correct}\:{then}\:{pl} \\ $$$${check}\:{the}\:{question}: \\ $$$${May}\:{be}\:{it}\:{is}\:{as}: \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{xz}+{z}^{\mathrm{2}} }+\sqrt{{z}^{\mathrm{2}} −\mathrm{6}{z}+\mathrm{9}}=\mathrm{4}? \\ $$$${In}\:{this}\:{case}\:{the}\:{solution}\:{will}\:{be} \\ $$$$\left(\mathrm{2},{z}\right)\:\&\:{z}=\mathrm{2}\:{or}\:\mathrm{3} \\ $$$${i}-{e}\:\left(\mathrm{2},\mathrm{2}\right)\:{or}\:\left(\mathrm{2},\mathrm{3}\right) \\ $$
Commented by mathdanisur last updated on 07/Jul/21
Sorry Ser, yes +4
$${Sorry}\:{Ser},\:{yes}\:+\mathrm{4} \\ $$

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