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Question Number 190339 by mr W last updated on 01/Apr/23
solve for x ∈R  (√(x−1))+(√(3x−5))+(√(4x−7))=4x−5
$${solve}\:{for}\:{x}\:\in\mathbb{R} \\ $$$$\sqrt{{x}−\mathrm{1}}+\sqrt{\mathrm{3}{x}−\mathrm{5}}+\sqrt{\mathrm{4}{x}−\mathrm{7}}=\mathrm{4}{x}−\mathrm{5} \\ $$
Answered by ajfour last updated on 01/Apr/23
(√(x−1))+(√(3x−5))+(√(4x−7))    =(1/2)(x−1+3x−5+4x−7)+(3/2)  ⇒  (p−1)^2 +(q−1)^2 +(r−1)^2 =0   p=(√(x−1))=q=(√(3x−5))=r=(√(4x−7))=1  ⇒ x=2 , 3x=6, 4x=8
$$\sqrt{{x}−\mathrm{1}}+\sqrt{\mathrm{3}{x}−\mathrm{5}}+\sqrt{\mathrm{4}{x}−\mathrm{7}} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{1}+\mathrm{3}{x}−\mathrm{5}+\mathrm{4}{x}−\mathrm{7}\right)+\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left({p}−\mathrm{1}\right)^{\mathrm{2}} +\left({q}−\mathrm{1}\right)^{\mathrm{2}} +\left({r}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\:{p}=\sqrt{{x}−\mathrm{1}}={q}=\sqrt{\mathrm{3}{x}−\mathrm{5}}={r}=\sqrt{\mathrm{4}{x}−\mathrm{7}}=\mathrm{1} \\ $$$$\Rightarrow\:{x}=\mathrm{2}\:,\:\mathrm{3}{x}=\mathrm{6},\:\mathrm{4}{x}=\mathrm{8} \\ $$
Commented by mehdee42 last updated on 01/Apr/23
very nice
$${very}\:{nice} \\ $$
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
my latest youtube video explaining  how long it takes to empty a water  filled tank, through a hole at its  bottom.
$${my}\:{latest}\:{youtube}\:{video}\:{explaining} \\ $$$${how}\:{long}\:{it}\:{takes}\:{to}\:{empty}\:{a}\:{water} \\ $$$${filled}\:{tank},\:{through}\:{a}\:{hole}\:{at}\:{its} \\ $$$${bottom}. \\ $$

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