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Monday, July 6, 2015

Problem of the Week

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.


  1. is 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

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

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