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

Hint: Here we will first assume the two roots of a quadratic equation as two variables. Then we will find the sum of the two roots using the property of the roots of the quadratic equation. We will then check the possible conditions for the roots as well as their sum to become prime numbers. Then we will solve the problem accordingly.

Complete step-by-step answer:
Let \[a{x^2} + bx + c = 0\] be the equation having prime numbers \[{p_1}\] ​and \[{p_2}\] ​as roots of this quadratic equation.
Now, we will find the sum of the roots which is equal to the negative of ratio of the coefficient of \[x\] to the coefficient of \[{x^2}\].
\[ \Rightarrow {p_1} + {p_2} = \dfrac{{ - b}}{a}\]
As \[{p_1}\] ​and \[{p_2}\] are prime numbers, this is only possible if \[a = 1\], therefore, we get
\[ \Rightarrow {p_1} + {p_2} = - b\]…….. \[\left( 1 \right)\]
Now, we will find the product of the roots which is equal to the ratio of the constant term to the coefficient of \[{x^2}\] of the quadratic equation.
\[ \Rightarrow {p_1} \times {p_2} = \dfrac{c}{a}\]
As \[{p_1}\] ​and \[{p_2}\] are prime numbers, this is only possible if \[a = 1\]
therefore, we get
\[ \Rightarrow {p_1} \times {p_2} = c\]…….. \[\left( 2 \right)\]
It is also given that the sum of the coefficients of the equation is a prime number.
Therefore, we get
\[ \Rightarrow a + b + c = {\rm{prime \,number}}\]
on substituting the value \[a = 1\], we get
\[ \Rightarrow 1 + b + c = {\rm{prime \,number}}\]
Now, we will put the value of \[b\] and \[c\] from equation 1 and equation 2, we get
\[ \Rightarrow 1 - \left( {{p_1} + {p_2}} \right) + {p_1} \times {p_2} = {\rm{prime\, number}}\]
On further simplification, we get
\[ \Rightarrow \left( {{p_1} - 1} \right)\left( {{p_2} - 1} \right) = {\rm{prime\, number}}\]
This is only possible when the whole term of one of the brackets is equal to 1 i.e.
\[\begin{array}{l} \Rightarrow {p_1} - 1 = 1\\ \Rightarrow {p_1} = 2\end{array}\]
And when \[\left( {{p_2} - 1} \right){\rm{ }} = 2\]
\[\begin{array}{l} \Rightarrow {p_2} - 1 = 2\\ \Rightarrow {p_2} = 3\end{array}\]
Thus, we have got the value of two roots. Now, we will find the sum of the roots
So,
\[ \Rightarrow {p_1} + {p_2} = 2 + 3 = 5\]
Hence, the only possible combination of prime roots is 2 and 3.
Thus, the correct option is option B.

Note: Here, we have seen that the roots of the quadratic equation are prime numbers and their sum is also a prime number. Prime number is defined as a number which is divisible by only two numbers i.e. 1 and the number itself. We need to remember that the number of roots of the polynomial is always equal to the highest power in the polynomial. The number of roots of a quadratic polynomial is 2 because the highest power in a quadratic polynomial is 2. Similarly the number of roots of cubic polynomials is 3 because the highest power in cubic polynomials is 3.