Question Number 190339 by mr W last updated on 01/Apr/23
$${solve}\:{for}\:{x}\:\in\mathbb{R} \\ $$$$\sqrt{{x}−\mathrm{1}}+\sqrt{\mathrm{3}{x}−\mathrm{5}}+\sqrt{\mathrm{4}{x}−\mathrm{7}}=\mathrm{4}{x}−\mathrm{5} \\ $$
Answered by ajfour last updated on 01/Apr/23
$$\sqrt{{x}−\mathrm{1}}+\sqrt{\mathrm{3}{x}−\mathrm{5}}+\sqrt{\mathrm{4}{x}−\mathrm{7}} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{1}+\mathrm{3}{x}−\mathrm{5}+\mathrm{4}{x}−\mathrm{7}\right)+\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left({p}−\mathrm{1}\right)^{\mathrm{2}} +\left({q}−\mathrm{1}\right)^{\mathrm{2}} +\left({r}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\:{p}=\sqrt{{x}−\mathrm{1}}={q}=\sqrt{\mathrm{3}{x}−\mathrm{5}}={r}=\sqrt{\mathrm{4}{x}−\mathrm{7}}=\mathrm{1} \\ $$$$\Rightarrow\:{x}=\mathrm{2}\:,\:\mathrm{3}{x}=\mathrm{6},\:\mathrm{4}{x}=\mathrm{8} \\ $$
Commented by mehdee42 last updated on 01/Apr/23
$${very}\:{nice} \\ $$
Commented by mr W last updated on 02/Apr/23
$$\:\cancel{ } \\ $$
Commented by ajfour last updated on 10/Apr/23
https://youtu.be/0I8a9urBLB8
Commented by ajfour last updated on 10/Apr/23
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