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

$$\mathrm{If}\mathrm{x},y\in \mathbb{R}satisfyinequation$$$$x-4\sqrt{y}=2\sqrt{x-y}.findrangeofx$$

Commented by MJS last updated on 23/Feb/20

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

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

$$thankyousir...$$

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

$$thankyouall$$

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

$$1)x⩾y$$$$2)y⩾0$$$$★nowify=0$$$$x=2\sqrt{x}$$$$\sqrt{x}(\sqrt{x}-2)=0$$$$soeitherx=0orx=4$$$$whenx=0then0y=0$$$$★whenx=y$$$$x-4\sqrt{x}=0$$$$\sqrt{x}(\sqrt{x}-4)=0$$$$x=0orx=16$$$$whenx=0theny=0$$$$letothersthrowlightin[thisproblem$$

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

$$y⩾0$$$$x⩾y⩾0$$$$x⩾4\sqrt{y}$$$$$$$$x^2-8x\sqrt{y}+16y=4(x-y)$$$$20y-8x\sqrt{y}+x^2-4x=0$$$$\sqrt{y}=\frac{2x\pm \sqrt{x(20-x)}}{10}⩾0$$$$x(20-x)⩾0$$$$\Rightarrow 0⩽x⩽20...\left(i\right)$$$$$$$$2x+\sqrt{x(20-x)}⩾0\Rightarrow alwaystrue$$$$2x-\sqrt{x(20-x)}⩾0$$$$4x^2⩾20x-x^2$$$$x^2⩾4x$$$$\Rightarrow x⩾4...\left(ii\right)$$$$from\left(i\right)and\left(ii\right)wegettherangeofx:$$$$4⩽x⩽20$$$$$$$$x^2-4(1+2\sqrt{y})x+20y=0$$$$x=2(1+2\sqrt{y})\pm 2\sqrt{1+4\sqrt{y}-y}$$$$1+4\sqrt{y}-y⩾0$$$$5-{(2-\sqrt{y})}^2⩾0$$$$\mid 2-\sqrt{y}\mid ⩽\sqrt{5}$$$$-\sqrt{5}⩽2-\sqrt{y}⩽\sqrt{5}$$$$\sqrt{y}⩽2+\sqrt{5}\Rightarrow y⩽{(2+\sqrt{5})}^2=9+4\sqrt{5}$$$$\sqrt{y}⩾2-\sqrt{5}\Rightarrow alwaystruewithy⩾0$$$$therangeofyis:$$$$0⩽y⩽9+4\sqrt{5}$$

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

$$thankyousir...$$

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