‘nother easy problem

October 17th, 2009 by ben Leave a reply »

People actually REPLY when I post easy problems! GASP!

(In this problem, \lfloor \cdot \rfloor represents the Greatest Integer function.)

If x=19941994, find the value of \left\lfloor \sqrt{x^2 - 10x + 29} \right\rfloor.

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2 comments

  1. Benji says:

    Oh, and no bruteforcing/calculators…

  2. Benji says:

    Solution, you lazy bums.

    [spoiler]

    Solution 1:

    Notice that the polynomial in the square root is four more than a perfect square.

    $latex \left\lfloor \sqrt{x^2 – 10x + 25 + 4} \right\rfloor$

    “Completing the square” you could call it.

    $latex \left\lfloor \sqrt{(x-5)^2 + 4} \right\rfloor$

    Since $latex x=19941994$, an additional 4 is negligible. Therefore, the answer is $latex 19941994 – 5 = \boxed{19941989}$

    Solution 2:

    If you’re too stupid to see the perfect square, just factor the polynomial using the quadratic formula. The roots are $latex 5 + 4i$ and $latex 5 – 4i$. Simply put them back in, and semi-expand it as a difference in squares.

    $latex \left\lfloor \sqrt{(x-5+4i)(x-5-4i)} \right\rfloor$

    And:

    $latex \left\lfloor \sqrt{(x-5)^2 + 4} \right\rfloor$

    [/spoiler]

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