There are a number of diff erent ways to approach this problem; the one I thought of uses "working backwards." I guessed, since the main complexities of x in the stated form are the radical signs, that x is of the form 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 diff erent 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