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Question Number 83420 by jagoll last updated on 02/Mar/20
find range x of function x−4(√y) = 2(√(x−y))
$$\mathrm{find}\:\mathrm{range}\:\mathrm{x}\:\mathrm{of}\:\mathrm{function}\: \\ $$$$\mathrm{x}−\mathrm{4}\sqrt{\mathrm{y}}\:=\:\mathrm{2}\sqrt{\mathrm{x}−\mathrm{y}} \\ $$
Commented by mathmax by abdo last updated on 02/Mar/20
we have 0≤y≤x (e)⇒(x−4(√y))^2 =4(x−y) ⇒ x^2 −8x(√y)+16y =4x−4y ⇒x^2 +16y−4x+4y =8x(√y) ⇒ x^2 −4x +20y =8x(√y) ⇒(x^2 −4x+20y)^2 =64x^2 y ⇒ (x^2 −4x)^2 +2(x^2 −4x)20y +400y^2 =64x^2 y ⇒ x^4 −8x^3 +16x^2 +40x^2 y−1600 xy +400y^2 −64 x^2 y =0 ⇒ 400y^2 +(40x^2 −64x^2 −1660x)y +x^4 −8x^3 +16x^2 =0 ⇒ 400y^2 −1684 y +x^4 −8x^3 +16x^2 =0 Δ^′ =(842)^2 −400(x^4 −8x^3 +16x^2 ) ⇒ y =((842+^− (√(842^2 −400x^4 +8.400x^3 −16.400x^2 )))/(400))...
$${we}\:{have}\:\mathrm{0}\leqslant{y}\leqslant{x}\:\:\:\left({e}\right)\Rightarrow\left({x}−\mathrm{4}\sqrt{{y}}\right)^{\mathrm{2}} =\mathrm{4}\left({x}−{y}\right)\:\Rightarrow \\ $$$${x}^{\mathrm{2}} −\mathrm{8}{x}\sqrt{{y}}+\mathrm{16}{y}\:=\mathrm{4}{x}−\mathrm{4}{y}\:\Rightarrow{x}^{\mathrm{2}} +\mathrm{16}{y}−\mathrm{4}{x}+\mathrm{4}{y}\:=\mathrm{8}{x}\sqrt{{y}}\:\Rightarrow \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}\:+\mathrm{20}{y}\:=\mathrm{8}{x}\sqrt{{y}}\:\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{20}{y}\right)^{\mathrm{2}} =\mathrm{64}{x}^{\mathrm{2}} {y}\:\Rightarrow \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)^{\mathrm{2}} +\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{4}{x}\right)\mathrm{20}{y}\:+\mathrm{400}{y}^{\mathrm{2}} \:=\mathrm{64}{x}^{\mathrm{2}} {y}\:\Rightarrow \\ $$$${x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} +\mathrm{16}{x}^{\mathrm{2}} +\mathrm{40}{x}^{\mathrm{2}} {y}−\mathrm{1600}\:{xy}\:+\mathrm{400}{y}^{\mathrm{2}} −\mathrm{64}\:{x}^{\mathrm{2}} {y}\:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{400}{y}^{\mathrm{2}} \:+\left(\mathrm{40}{x}^{\mathrm{2}} −\mathrm{64}{x}^{\mathrm{2}} −\mathrm{1660}{x}\right){y}\:+{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} +\mathrm{16}{x}^{\mathrm{2}} \:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{400}{y}^{\mathrm{2}} −\mathrm{1684}\:{y}\:+{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} \:+\mathrm{16}{x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta^{'} =\left(\mathrm{842}\right)^{\mathrm{2}} −\mathrm{400}\left({x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} \:+\mathrm{16}{x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${y}\:=\frac{\mathrm{842}\overset{−} {+}\sqrt{\mathrm{842}^{\mathrm{2}} −\mathrm{400}{x}^{\mathrm{4}} +\mathrm{8}.\mathrm{400}{x}^{\mathrm{3}} −\mathrm{16}.\mathrm{400}{x}^{\mathrm{2}} }}{\mathrm{400}}… \\ $$
Commented by mr W last updated on 02/Mar/20
to jagoll sir: it′s wrong! x^2 −(4t+16)x+20t^2 = 0 is not a quadratic equation for x! because t contains x!
$${to}\:{jagoll}\:{sir}: \\ $$$${it}'{s}\:{wrong}! \\ $$$$\mathrm{x}^{\mathrm{2}} −\left(\mathrm{4t}+\mathrm{16}\right)\mathrm{x}+\mathrm{20t}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$${is}\:{not}\:{a}\:{quadratic}\:{equation}\:{for}\:{x}! \\ $$$${because}\:{t}\:{contains}\:{x}! \\ $$
Commented by MJS last updated on 02/Mar/20
x^2 −(4t+16)x+20t^2 =0 but t=(√(x−y)) ⇒ this is not a polynome of 2^(nd) degree in x 1−(√5)≤t≤1+(√5) 0≤(√(x−y))≤1+(√5) 0≤x−y≤6+2(√5) and now you know nothing about x
$${x}^{\mathrm{2}} −\left(\mathrm{4}{t}+\mathrm{16}\right){x}+\mathrm{20}{t}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{but}\:{t}=\sqrt{{x}−{y}}\:\Rightarrow\:\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of} \\ $$$$\mathrm{2}^{\mathrm{nd}} \:\mathrm{degree}\:\mathrm{in}\:{x} \\ $$$$\mathrm{1}−\sqrt{\mathrm{5}}\leqslant{t}\leqslant\mathrm{1}+\sqrt{\mathrm{5}} \\ $$$$\mathrm{0}\leqslant\sqrt{{x}−{y}}\leqslant\mathrm{1}+\sqrt{\mathrm{5}} \\ $$$$\mathrm{0}\leqslant{x}−{y}\leqslant\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\mathrm{and}\:\mathrm{now}\:\mathrm{you}\:\mathrm{know}\:\mathrm{nothing}\:\mathrm{about}\:{x} \\ $$
Commented by jagoll last updated on 02/Mar/20
oo yes. i understand now. thank you mister w and mister mjs
$$\mathrm{oo}\:\mathrm{yes}.\:\mathrm{i}\:\mathrm{understand}\:\mathrm{now}.\:\mathrm{thank}\:\mathrm{you} \\ $$$$\mathrm{mister}\:\mathrm{w}\:\mathrm{and}\:\mathrm{mister}\:\mathrm{mjs} \\ $$
Answered by mr W last updated on 02/Mar/20
we have y≥0 x≥y≥0 x≥4(√y) let (√y)=v≥0 x−4v=2(√(x−v^2 )) x^2 −8xv+16v^2 =4(x−v^2 ) 20v^2 −8xv+x^2 −4x=0 ⇒v=((2x±(√(x(20−x))))/(10)) x(20−x)≥0 ⇒0≤x≤20 ...(i) v≥0 ⇒2x+(√(x(20−x)))≥0 ←always true! ⇒2x−(√(x(20−x)))≥0 ⇒5x(x−4)≥0 ⇒x≥4 ...(ii) from (i) and (ii) we get range of x: 4≤x≤20 x^2 −4(1+2v)x+20v^2 =0 x=2(1+2v)±2(√(1+4v−v^2 ))>0 1+4v−v^2 ≥0 v^2 −4v−1≤0 ⇒2−(√5)≤v≤2+(√5) since v≥0, ⇒0≤v≤2+(√5) ⇒0≤y≤(2+(√5))^2 =9+4(√5) ...(iii) from (iii) we get range of y: 0≤y≤9+4(√5)
$${we}\:{have} \\ $$$${y}\geqslant\mathrm{0} \\ $$$${x}\geqslant{y}\geqslant\mathrm{0} \\ $$$${x}\geqslant\mathrm{4}\sqrt{{y}} \\ $$$$ \\ $$$${let}\:\sqrt{{y}}={v}\geqslant\mathrm{0} \\ $$$${x}−\mathrm{4}{v}=\mathrm{2}\sqrt{{x}−{v}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} −\mathrm{8}{xv}+\mathrm{16}{v}^{\mathrm{2}} =\mathrm{4}\left({x}−{v}^{\mathrm{2}} \right) \\ $$$$\mathrm{20}{v}^{\mathrm{2}} −\mathrm{8}{xv}+{x}^{\mathrm{2}} −\mathrm{4}{x}=\mathrm{0} \\ $$$$\Rightarrow{v}=\frac{\mathrm{2}{x}\pm\sqrt{{x}\left(\mathrm{20}−{x}\right)}}{\mathrm{10}} \\ $$$${x}\left(\mathrm{20}−{x}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}\leqslant{x}\leqslant\mathrm{20}\:\:\:\:\:\:…\left({i}\right) \\ $$$${v}\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{x}+\sqrt{{x}\left(\mathrm{20}−{x}\right)}\geqslant\mathrm{0}\:\leftarrow{always}\:{true}! \\ $$$$\Rightarrow\mathrm{2}{x}−\sqrt{{x}\left(\mathrm{20}−{x}\right)}\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{5}{x}\left({x}−\mathrm{4}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow{x}\geqslant\mathrm{4}\:\:\:…\left({ii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get}\:{range}\:{of}\:{x}: \\ $$$$\mathrm{4}\leqslant{x}\leqslant\mathrm{20} \\ $$$$ \\ $$$${x}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}+\mathrm{2}{v}\right){x}+\mathrm{20}{v}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\mathrm{2}\left(\mathrm{1}+\mathrm{2}{v}\right)\pm\mathrm{2}\sqrt{\mathrm{1}+\mathrm{4}{v}−{v}^{\mathrm{2}} }>\mathrm{0} \\ $$$$\mathrm{1}+\mathrm{4}{v}−{v}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${v}^{\mathrm{2}} −\mathrm{4}{v}−\mathrm{1}\leqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}−\sqrt{\mathrm{5}}\leqslant{v}\leqslant\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$${since}\:{v}\geqslant\mathrm{0}, \\ $$$$\Rightarrow\mathrm{0}\leqslant{v}\leqslant\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{0}\leqslant{y}\leqslant\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{9}+\mathrm{4}\sqrt{\mathrm{5}}\:\:\:…\left({iii}\right) \\ $$$${from}\:\left({iii}\right)\:{we}\:{get}\:{range}\:{of}\:{y}: \\ $$$$\mathrm{0}\leqslant{y}\leqslant\mathrm{9}+\mathrm{4}\sqrt{\mathrm{5}} \\ $$

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