Prime Roots

For a quadratic equation of the form $ax^2 + bx + c = 0$ having roots $\alpha$ and $\beta$ , $\alpha + \beta = \frac{-b}{a} = 12$

The only prime values of $\alpha$ and $\beta$ that satisfy this equation are $7$ and $5$ .

We also know that, $\alpha \times \beta = \frac{c}{a} = k$

So, the only value for $k$ is $\boxed{35}$

We shall start by determining its roots.

Using the quadratic formula (which is $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ ), we have $\frac{12 \pm \sqrt{144 - 4k}}{2}$ as roots. On further simplifying, we have

$\frac{12}{2} \pm \sqrt{\frac{144 - 4k}{4}}$ which is $6 \pm \sqrt{36-k}$ .

The only value that satisfies $6 \pm a =$ prime number is $a = 1$ .

Putting $k = 35$ in $6 \pm \sqrt{36-k}$ , we get the prime numbers $5$ and $7$ .

Therefore, the sum of the values of $k$ is $\boxed{35}$ .

Hey yo,

as x^2 - 12x + k = 0, as we all know, prime numbers = [2,3,5,7,11,13....],

as x^2 - 12x + k =0 -----> to get k must be always ( x-a)(x-b), where ab = k,

by logical thinking,

as for a and b that is determined by 12(only 5 and 7),

a = 5, b = 7 ------> ab = 35.... (x-5)(x-7) = x^2 - 12x + k....

haha....just looking for fun.....