*a*+

*b**sqrt(

*c*) for some rational numbers

*a*and

*b*and some positive integer

*c*that isn't a perfect square. Then,

Since 3 is the only quantity in the radical in the RHS, I surmised that

*c = 3*, and thus

*2ab = 4*. Because

*a*,

*b*, and

*c*are all rational, neither

*a^2*nor

*(b^2)c*can contribute to the sqrt(3) term in the RHS. Then, equating the rational terms, I solved for

*a*using the quadratic formula:

As I required a to be rational, a = +/- 2, and thus b = +/- 1 from (5), with the signs on each the same. So, since the convention is to take the positive square root,

Did you think of a different solution? What did you think of my method? Let us know in the comments, and be sure to check out the rest of our Problems of the Week.

I like it

ReplyDeleteis too long this way.but you can write 7 as 4+3 and being squares of 2 and sqrt(3) then following (a+b)^2

ReplyDeleteRadu, are you talking about completing the square on 3 + 4 sqrt (3) when you say following (a + b)^2 ?

ReplyDeleteI like Radu's method: x^2=4+4sqrt(3)+sqrt(3)=(2+sqrt(3))^2

ReplyDelete