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
·
10 months, 4 weeks ago

Log in to reply

@Aareyan Manzoor
–
I think that there is one more solution pair, namely \((x,y) = (45,180)\), as

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

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
·
10 months, 4 weeks ago

Log in to reply

pls help me.If w is one of the root of x2+x+2.find w10+w5+3
–
Shalom Will
·
10 months, 2 weeks ago

Log in to reply

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
·
10 months, 3 weeks ago

## Comments

Sort by:

TopNewestok. 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 · 10 months, 4 weeks ago

Log in to reply

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

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 · 10 months, 4 weeks ago

Log in to reply

pls help me.If w is one of the root of x2+x+2.find w10+w5+3 – Shalom Will · 10 months, 2 weeks ago

Log in to reply

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 · 10 months, 3 weeks ago

Log in to reply