If the roots of the equation are integers then find the value of $k$.

Question

The question says roots of $x^2-kx+36=0$ are integers then the number of values for $k$ are?

I know roots are integral if discriminant is a perfect square of an integer, but using this I get infinite values for $k$. Which values should I reject?

Answer 1

HINT:

Let $k^2-4\cdot1\cdot36=y^2\iff(k+y)(k-y)=144$

As $k+y\pm(k-y)$ are even, $k-y,k+y$ have the same parity, hence both must be even

$$\dfrac{k+y}2\cdot\dfrac{k-y}2=\dfrac{144}4=?$$

What are the positive divisors of $36?$

Also as $k+y\ge k-y,36=\dfrac{k+y}2\cdot\dfrac{k-y}2\le\dfrac{(k+y)^2}4$

$\implies k+y\ge\sqrt{4\cdot36}$ as $k+y>0$

$\implies\dfrac{k+y}2\ge6$ and must divide $36$

Answer 2

The product of the roots is $36$, and the sum is $k$. There are $5$ ways to decompose $36$ as a product of positive factors, if order does not count, and an equal number of ways to decompose $36$ as a product of negative factors.

Thus there are $10$ possible values of $k$.

Answer 3

Uh! I remember solving very similar problem with a different equation. Let's say I will give you an explanation for that, and you could figured out. So, if the roots of the equation $x^2 + px + 36 = 0$ are integers, how many values of 'p' are possible?

Looks simple but quite interesting, definitely the answer is not 2.

Given the quadratic equation: $$x^2 + px + 36 = 0$$

We know that the sum of the roots, denoted as $-p$, is related to the product of the roots, which is $36$.

Now, since $36 = 2^2 \cdot 3^2$, the number of factors of $36$ is given by $(2 + 1)(2 + 1) = 9$.

Therefore, the number of ways in which $36$ can be expressed as the product of two distinct natural numbers $=\frac{9-1}{2}=4(1 \times 36=2 \times 18=3 \times 12 = 4 \times 9)$. Additionally, there's one way to express 36 as a product of the same two numbers, which is $6 \times 6$.

Hence, there are a total of $5$ different values of 'p' that make the roots of the equation integers. These values are obtained by considering positive and including negative pairs: $(-1 \times -36= -2 \times -18= -3 \times -12= -4 \times-9)$.

Therefore total of $10$ different values of p are possible.