Given the equation $x^2-14x+m=0$, where m is a positive integer. If $p_1,p_2$ are roots, then find the value of $K=(p_1+p_2)^2+2p_1p_2$

Question

Given the equation $x^2-14x+m=0$, where $m$ is a positive integer. If roots $p_1, p_2$ are distinct, positive, prime integers, then find the value of $K=(p_1+p_2)^2+2p_1p_2$

I tried to solve this question using the Vieta formulas and said $p_1+p_2=14$ and $p_1p_2=m$. But I got stuck there, I assume that you have to work out what the value of $m$ is, but I can't manage to work it out. Could you please explain to me how to solve this question?

Answer 1

Well, if p1 and p2 are distinct, then there is only a single pair of positive primes that satisfies the condition $p_1+p_2=14$, and these are $3$ and $11$.

Else, if $p_1$ and $p_2$ are not distinct, the question could have two possible answers, because both $7$ and $7$ as well as $3$ and $11$ satisfy the equation.

Can you go ahead now?

Answer 2

Well the roots of $x^2 - 14x + m = (x^2 -14x +49) + (m-49) = (x-7)^2 -(49-m)$ are $7 \pm \sqrt{49-m}$.

So $p_1+ p_2 = 14$ and $p_1p_2 = 49 - (49-m) = m$.

So $m$ is the product of two distinct prime so that $49 - m$ is perfect square. Well, quit worrying and start scurrying.

The perfect squares less than $49$ are $1,4,9, 16, 25, 36$ so $m = 48,45, 40, 33,24,13$ and of those the only one that is product of two distinct primes are $33=3\times 11$

So $(p_1 + p_1)^2 + 2p_1p_2 = 14^2 + 2\cdot 33$.

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Argh a dope slap to me for getting $49-m$ must be a perfect square but not getting that $14$ is the sum of distinct primes. It is still trial and error, but $1+13,2+12,3+11,4+10, 5+9, 8+6, 7+7$ so $3,11$ are the only sum of two primes is an easier trial by error.