Question

A quadratic equation with integral coefficients has two different prime numbers as its roots. If the sum of the coefficients of the equation is prime, then the sum of the roots is

• A. $2$
• B. $5$
• C. $7$
• D. $11$

Answer 1

The correct option is B $5$

Let $a{x}^2+bx+c=0$ be the equation having prime numbers $p_1,p_2$ as roots.
Since, the roots of the equation are integers. We can write the equation as :
$a(x-p_1)(x-p_2)=0$
Now, the sum of the coefficients is a prime number. In the above equation the sum of the coefficients can be written as $a\times \left(k\right)$, where $k$ is the sum of the coefficients of the equation $(x-p_1)(x-p_2)=0$.
This means, either $a=1$ or the sum of coefficients of $(x-p_1)(x-p_2)=0$ is zero. The second case is not possible as it would mean $x=1$ is a solution of the equation which is not prime.
Hence, $a=1$
The equation is
$x^2+bx+c=0$
The sum of the coefficients $=1+b+c$
There are two possible cases - 1) both roots are odd 2)one root is odd and the other one is even (2)
Case 1 : Both the prime numbers are odd
The equation can be written as $x^2-(p_1+p_2)x+\left(p_1{p}_2\right)=0$
Now the sum of the coefficients $=1-(p_1+p_2)+p_1{p}_2$
This is an even number and prime. Hence, the only possibility is the sum of the roots $=2$
$\Rightarrow 1-p_1-p_2+p_1{p}_2=2$
$\Rightarrow p_1=\frac{p_2+1}{p_2-1}$
$p_1$ cannot be an integer unless, $p_2=3$ ,which is not possible as $p_1$ also has to be a prime number.
Hence, both the roots cannot be odd.
Case 2 - One root is odd and the other is even.
As the roots are prime, $p_1=2$ .
Sum of the co-efficients $=1-(2+p_2)+2p_2$
This will be an even number and it is prime. Hence, the sum of the roots must be $2$
$\Rightarrow 1-2-p_2+2p_2=2$
$\Rightarrow p_2=3$
Hence, the only possible combination of prime roots is $2$ and $3$
So, option B is correct.