Math Is Fun Forum

Agnishom

Real Member

From: Riemann Sphere

Registered: 2011-01-29

Posts: 25,002

Website

Sum of roots

How many ordered pairs of non-negative integers (a,b) are there such that √a+√b=√432 ?

Since, √432 = 12√3, therefore the pairs are 0  + (12√3)^2, (√3)^2 + (11√3)^2, (2√3)^2 + (9√3)^2, and so on. So a total of 12 ordered pairs.

Why is this wrong?

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Nehushtan

Member

Registered: 2013-03-09

Posts: 957

Re: Sum of roots

Since, √432 = 12√3, therefore the pairs are 0  + (12√3)^2, (√3)^2 + (11√3)^2, (2√3)^2 + (9√3)^2, and so on. So a total of 12 ordered pairs.

Won’t there actually be 13 ordered pairs (because of the 0)?

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bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Sum of roots

I am getting 6 with a>b.

13 with order counting.

I counted them up.

{0,432}
{3,363}
{12,300}
{27,243}
{48,192}
{75,147}
{108,108}
{147,75}
{192,48}
{243,27}
{300,12}
{363,3}
{432,0}

Oh, sorry it already got answered.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Agnishom

Real Member

From: Riemann Sphere

Registered: 2011-01-29

Posts: 25,002

Website

Re: Sum of roots

Oh yeah! I missed the question totally

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'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.