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if-x-y-1-2-y-x-and-2x-2-7-thin-faind-volue-of-x-y-




Question Number 172703 by mathlove last updated on 30/Jun/22
if (√(x/y))+(1/( (√2)))=(√(y/x)) and 2x^2 =7  thin faind volue of x∙y=?
$${if}\:\sqrt{\frac{{x}}{{y}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\sqrt{\frac{{y}}{{x}}}\:{and}\:\mathrm{2}{x}^{\mathrm{2}} =\mathrm{7} \\ $$$${thin}\:{faind}\:{volue}\:{of}\:{x}\centerdot{y}=? \\ $$
Answered by Rasheed.Sindhi last updated on 30/Jun/22
(√(x/y))+(1/( (√2)))=(√(y/x)) and 2x^2 =7 ; xy=?  Let (√(x/y)) =a≥0  (1/a)−a=(1/( (√2) ))  (√2) (1−a^2 )=a  (√2) a^2 +a−(√2) =0  a=((−1±(√(1+8)))/(2(√2)))=(1/( (√2) ))^✓ ,  ((−2)/( (√2)))(rejected)  (√(x/y)) =(1/( (√2) ))⇒(x/y)=(1/2)⇒y=2x             ⇒xy=x.2x=2x^2 =7
$$\sqrt{\frac{{x}}{{y}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\sqrt{\frac{{y}}{{x}}}\:{and}\:\mathrm{2}{x}^{\mathrm{2}} =\mathrm{7}\:;\:{xy}=? \\ $$$${Let}\:\sqrt{\frac{{x}}{{y}}}\:={a}\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{a}}−{a}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:} \\ $$$$\sqrt{\mathrm{2}}\:\left(\mathrm{1}−{a}^{\mathrm{2}} \right)={a} \\ $$$$\sqrt{\mathrm{2}}\:{a}^{\mathrm{2}} +{a}−\sqrt{\mathrm{2}}\:=\mathrm{0} \\ $$$${a}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{8}}}{\mathrm{2}\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\:^{\checkmark} ,\:\:\frac{−\mathrm{2}}{\:\sqrt{\mathrm{2}}}\left({rejected}\right) \\ $$$$\sqrt{\frac{{x}}{{y}}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{y}=\mathrm{2}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{xy}={x}.\mathrm{2}{x}=\mathrm{2}{x}^{\mathrm{2}} =\mathrm{7} \\ $$
Commented by mathlove last updated on 30/Jun/22
thanks
$${thanks} \\ $$

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