×

# Help me in this question

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

Note by Ved Sharda
1 year, 10 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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 $(k-1)(k_2-1)20=20\Longrightarrow k=2,k_1=2\Longrightarrow x=80,y=80$

- 1 year, 10 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.

- 1 year, 10 months ago

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

- 1 year, 10 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

- 1 year, 10 months ago