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Question Number 82719 by john santu last updated on 23/Feb/20

If x,y ∈R satisfy in equation x−4(√y) = 2(√(x−y)) . find range of x

$$\mathrm{If}\:\mathrm{x},{y}\:\in\mathbb{R}\:{satisfy}\:{in}\:{equation}\: \\ $$$${x}−\mathrm{4}\sqrt{{y}}\:=\:\mathrm{2}\sqrt{{x}−{y}}\:.\:{find}\:{range}\:{of}\:{x} \\ $$

Commented by MJS last updated on 23/Feb/20

4≤x≤20 I will show later for 4≤x<16 we get 1 solution for y for 16≤x≤20 we get 2 solutions for y

$$\mathrm{4}\leqslant{x}\leqslant\mathrm{20} \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{show}\:\mathrm{later} \\ $$$$\mathrm{for}\:\mathrm{4}\leqslant{x}<\mathrm{16}\:\mathrm{we}\:\mathrm{get}\:\mathrm{1}\:\mathrm{solution}\:\mathrm{for}\:{y} \\ $$$$\mathrm{for}\:\mathrm{16}\leqslant{x}\leqslant\mathrm{20}\:\mathrm{we}\:\mathrm{get}\:\mathrm{2}\:\mathrm{solutions}\:\mathrm{for}\:{y} \\ $$

Commented by TANMAY PANACEA last updated on 23/Feb/20

thank you sir...

$${thank}\:{you}\:{sir}... \\ $$

Commented by john santu last updated on 24/Feb/20

thank you all

$${thank}\:{you}\:{all} \\ $$

Answered by TANMAY PANACEA last updated on 23/Feb/20

1)x≥y 2)y≥0 ★now if y=0 x=2(√x) (√x) ((√x) −2)=0 so either x=0 or x=4 when x=0 then0 y=0 ★when x=y x−4(√x) =0 (√x) ((√x) −4)=0 x=0 or x=16 when x=0 then y=0 let others throw light in[this problem

$$\left.\mathrm{1}\right){x}\geqslant{y} \\ $$$$\left.\mathrm{2}\right){y}\geqslant\mathrm{0} \\ $$$$\bigstar{now}\:{if}\:{y}=\mathrm{0} \\ $$$${x}=\mathrm{2}\sqrt{{x}}\: \\ $$$$\sqrt{{x}}\:\left(\sqrt{{x}}\:−\mathrm{2}\right)=\mathrm{0} \\ $$$${so}\:{either}\:{x}=\mathrm{0}\:\:\:{or}\:{x}=\mathrm{4} \\ $$$${when}\:{x}=\mathrm{0}\:{then}\mathrm{0}\:\:{y}=\mathrm{0} \\ $$$$\bigstar{when}\:{x}={y} \\ $$$${x}−\mathrm{4}\sqrt{{x}}\:=\mathrm{0} \\ $$$$\sqrt{{x}}\:\left(\sqrt{{x}}\:−\mathrm{4}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\:\:{or}\:{x}=\mathrm{16} \\ $$$${when}\:{x}=\mathrm{0}\:\:{then}\:{y}=\mathrm{0} \\ $$$${let}\:{others}\:{throw}\:{light}\:{in}\left[{this}\:{problem}\right. \\ $$

Answered by mr W last updated on 23/Feb/20

y≥0 x≥y≥0 x≥4(√y) x^2 −8x(√y)+16y=4(x−y) 20y−8x(√y)+x^2 −4x=0 (√y)=((2x±(√(x(20−x))))/(10))≥0 x(20−x)≥0 ⇒0≤x≤20 ...(i) 2x+(√(x(20−x)))≥0 ⇒always true 2x−(√(x(20−x)))≥0 4x^2 ≥20x−x^2 x^2 ≥4x ⇒x≥4 ...(ii) from (i) and (ii)we get the range of x: 4≤x≤20 x^2 −4(1+2(√y))x+20y=0 x=2(1+2(√y))±2(√(1+4(√y)−y)) 1+4(√y)−y≥0 5−(2−(√y))^2 ≥0 ∣2−(√y)∣≤(√5) −(√5)≤2−(√y)≤(√5) (√y)≤2+(√5) ⇒y≤(2+(√5))^2 =9+4(√5) (√y)≥2−(√5) ⇒always true with y≥0 the range of y is: 0≤y≤9+4(√5)

$${y}\geqslant\mathrm{0} \\ $$$${x}\geqslant{y}\geqslant\mathrm{0} \\ $$$${x}\geqslant\mathrm{4}\sqrt{{y}} \\ $$$$ \\ $$$${x}^{\mathrm{2}} −\mathrm{8}{x}\sqrt{{y}}+\mathrm{16}{y}=\mathrm{4}\left({x}−{y}\right) \\ $$$$\mathrm{20}{y}−\mathrm{8}{x}\sqrt{{y}}+{x}^{\mathrm{2}} −\mathrm{4}{x}=\mathrm{0} \\ $$$$\sqrt{{y}}=\frac{\mathrm{2}{x}\pm\sqrt{{x}\left(\mathrm{20}−{x}\right)}}{\mathrm{10}}\geqslant\mathrm{0} \\ $$$${x}\left(\mathrm{20}−{x}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}\leqslant{x}\leqslant\mathrm{20}\:\:\:...\left({i}\right) \\ $$$$ \\ $$$$\mathrm{2}{x}+\sqrt{{x}\left(\mathrm{20}−{x}\right)}\geqslant\mathrm{0}\:\Rightarrow{always}\:{true} \\ $$$$\mathrm{2}{x}−\sqrt{{x}\left(\mathrm{20}−{x}\right)}\geqslant\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \geqslant\mathrm{20}{x}−{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \geqslant\mathrm{4}{x} \\ $$$$\Rightarrow{x}\geqslant\mathrm{4}\:\:\:...\left({ii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right){we}\:{get}\:{the}\:{range}\:{of}\:{x}: \\ $$$$\mathrm{4}\leqslant{x}\leqslant\mathrm{20} \\ $$$$ \\ $$$${x}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}+\mathrm{2}\sqrt{{y}}\right){x}+\mathrm{20}{y}=\mathrm{0} \\ $$$${x}=\mathrm{2}\left(\mathrm{1}+\mathrm{2}\sqrt{{y}}\right)\pm\mathrm{2}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{{y}}−{y}} \\ $$$$\mathrm{1}+\mathrm{4}\sqrt{{y}}−{y}\geqslant\mathrm{0} \\ $$$$\mathrm{5}−\left(\mathrm{2}−\sqrt{{y}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\mid\mathrm{2}−\sqrt{{y}}\mid\leqslant\sqrt{\mathrm{5}} \\ $$$$−\sqrt{\mathrm{5}}\leqslant\mathrm{2}−\sqrt{{y}}\leqslant\sqrt{\mathrm{5}} \\ $$$$\sqrt{{y}}\leqslant\mathrm{2}+\sqrt{\mathrm{5}}\:\Rightarrow{y}\leqslant\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{9}+\mathrm{4}\sqrt{\mathrm{5}} \\ $$$$\sqrt{{y}}\geqslant\mathrm{2}−\sqrt{\mathrm{5}}\:\Rightarrow{always}\:{true}\:{with}\:{y}\geqslant\mathrm{0} \\ $$$${the}\:{range}\:{of}\:{y}\:{is}: \\ $$$$\mathrm{0}\leqslant{y}\leqslant\mathrm{9}+\mathrm{4}\sqrt{\mathrm{5}} \\ $$

Commented by TANMAY PANACEA last updated on 23/Feb/20

thank you sir...

$${thank}\:{you}\:{sir}... \\ $$

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