Quadratic equation

Algebra Level 5

$\displaystyle nx^2+2nx+n-9=0$

Let $S$ the sum of the possible values of $n$ so that the equation above has integer solution.

Find the value of $\lfloor{S}\rfloor$ .

Note : $n$ may not be an integer.

The answer is 14.

2 solutions

Jake Lai
Oct 31, 2015

Rearrange the given equation to get

$n(x+1)^2 = 9$

Hence, we know that $n = \frac{9}{(x+1)^2}$ . $(x+1)^2$ can take on the values $(x+1)^2 = 1,4,9,\ldots$ so we obtain

$S = \sum_{k=1}^{\infty} \frac{9}{k^2} = 9 \times \frac{\pi^2}{6} \approx 14.804$

and so $\lfloor S \rfloor = \boxed{14}$ .

Moderator note:

Good approach. For completeness, you should point out that we get distinct values of $n$ , for each chosen value of $(x+1) ^2$ . If we were counting by value of $x$ , then we would have twice the sum.

Chan Lye Lee
Oct 9, 2015

Using the quadratic formula, $\displaystyle x = \frac{-2n \pm \sqrt{4n^2-(4n(n-9))}}{2n} = -1 \pm \frac{3}{\sqrt{n}}$ . If $\displaystyle x \in \mathbb{Z}$ , then $\displaystyle \frac{3}{\sqrt{n}}=k$ for some integer $k$ . Hence, $\displaystyle n=\frac{9}{k^2}$ and thus $\displaystyle S = \sum_{k=1}^{\infty} \frac{9}{k^2} = \frac{9 \pi ^2}{6}$ . Now, $\displaystyle \lfloor S \rfloor = 14$ .

how did you get summation of 9/k^2? especially how did term of pi emerge?

Sarthak Singla - 3 years, 11 months ago

Its simply reimann zeta function

That is summation of $\dfrac{1}{1^2}+\dfrac{1}{2^2}+...\infty= \dfrac{\pi^2}{6}$

Ayaen Shukla - 3 years, 9 months ago

Quadratic equation

The answer is 14.

2 solutions

Moderator note:

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