# ‘nother easy problem

October 17th, 2009 by ben

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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