Help me in this question

Find all positive integers $x$ and $y$ satisfying $\dfrac1{\sqrt x} + \dfrac1{\sqrt y} = \dfrac1{\sqrt{20}}$ .

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Comments

ok. multiply both sides by $\sqrt{20xy}$ $\sqrt{20}(\sqrt{x}+\sqrt{y})=\sqrt{xy}$ change $(\sqrt{x},\sqrt{y})\to (a,b)$ $ab-\sqrt{20}(a+b)+20=20\\ (a-\sqrt{20})(b-\sqrt{20})=20$ notice how only $a=k\sqrt{20},b=k_2\sqrt{20}$ would yield positive integer answers. we check $(k-1)(k_2-1)20=20\Longrightarrow k=2,k_1=2\Longrightarrow x=80,y=80$

Aareyan Manzoor - 5 years, 5 months ago

I think that there is one more solution pair, namely $(x,y) = (45,180)$ , as

$\dfrac{1}{\sqrt{45}} + \dfrac{1}{\sqrt{180}} = \dfrac{1}{3\sqrt{5}} + \dfrac{1}{6\sqrt{5}} = \left(\dfrac{1}{3} + \dfrac{1}{6}\right)\dfrac{1}{\sqrt{5}} = \dfrac{1}{2\sqrt{5}} = \dfrac{1}{\sqrt{20}}.$

My approach was to note that as $\sqrt{20} = 2\sqrt{5}$ we must have $x = 5m^{2}$ and $y = 5n^{2}$ for some positive integers $m,n$ both greater than $2$ , yielding the equation

$\dfrac{1}{m} + \dfrac{1}{n} = \dfrac{1}{2} \Longrightarrow 2m + 2n = mn \Longrightarrow mn - 2m - 2n = 0 \Longrightarrow (m - 2)(n - 2) = 4.$

As $4 = 2*2 = 1*4$ we can either have $(m,n) = (4,4)$ or $(m,n) = (3,6)$ , the former of which yields your solution and the latter my additional solution.

Brian Charlesworth - 5 years, 5 months ago

Solve for x: 1/sqrt(x)+1/sqrt(y) = 1/(2 sqrt(5)) 1/(2 sqrt(5)) = 1/(2 sqrt(5)): 1/sqrt(x)+1/sqrt(y) = 1/(2 sqrt(5)) Bring 1/sqrt(x)+1/sqrt(y) together using the common denominator sqrt(x) sqrt(y): (sqrt(x)+sqrt(y))/(sqrt(x) sqrt(y)) = 1/(2 sqrt(5)) Cross multiply: 2 sqrt(5) (sqrt(x)+sqrt(y)) = sqrt(x) sqrt(y) 2 sqrt(5) (sqrt(x)+sqrt(y)) = sqrt(x) sqrt(y) is equivalent to sqrt(x) sqrt(y) = 2 sqrt(5) (sqrt(x)+sqrt(y)): sqrt(x) sqrt(y) = 2 sqrt(5) (sqrt(x)+sqrt(y)) Subtract 2 sqrt(5) (sqrt(x)+sqrt(y)) from both sides: sqrt(x) sqrt(y)-2 sqrt(5) (sqrt(x)+sqrt(y)) = 0 sqrt(x) sqrt(y)-2 sqrt(5) (sqrt(x)+sqrt(y)) = sqrt(x) (sqrt(y)-2 sqrt(5))-2 sqrt(5) sqrt(y): sqrt(x) (sqrt(y)-2 sqrt(5))-2 sqrt(5) sqrt(y) = 0 Add 2 sqrt(5) sqrt(y) to both sides: sqrt(x) (sqrt(y)-2 sqrt(5)) = 2 sqrt(5) sqrt(y) Divide both sides by sqrt(y)-2 sqrt(5): sqrt(x) = (2 sqrt(5) sqrt(y))/(sqrt(y)-2 sqrt(5)) Raise both sides to the power of two: Answer: x = (20 y)/(sqrt(y)-2 sqrt(5))^2

sharad yadav - 5 years, 4 months ago

pls help me.If w is one of the root of x2+x+2.find w10+w5+3

shalom will - 5 years, 4 months ago

find the value of x/y if (3/√y) - (1/√x) = 2/(√x + √y)

tim oneil - 3 years, 5 months ago

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`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$