Prime Quadratics

$\text{As 1 is a root of the equation,therefore,}$ $a+b-2c+1=0$ . $\text{which is equivalent to}$ $\dfrac{a+b+1}2=c$ .

$\text{As c is a natural prime number,it is clear that a+b must be odd.But a and b are also primes.}$ $\textit{All primes except 2 are odd.}$

$\text{And the sum of 2 numbers can be odd if and only if they both have different parity.So one is odd,other is even.}$

$\text{This implies that if a is an odd prime,then b is necessarily 2,and if b is an odd prime,then a is necessarily 2.}$

$\text{So,either a or b is 2,but not both.This eliminates the first 3 options.So the answer must be option d.)}$ .

Let,

$a x^2 + bx - 2c + 1 = \left( x-1 \right) \left(ax + k \right)$

Expanding the RHS of this equation, we get,

$\begin{aligned} a x^2 + bx - 2c + 1 &= a x^2 + kx - ax - k \\ & = a x^2 + \left(k-a \right) x - k \end{aligned}$

Comparing terms on both sides, we obtain,

$k = 2c - 1 \ \cdots(1) \\ k = a + b \ \cdots(2)$

Consider the equation $2b + 3c = 25$ , here,

$c = \dfrac{25- 2b}{3}$

Substituting the value of $c$ in $(1)$ gives,

$k = \dfrac{47 - 4b}{3}$

Equating this value of $k$ with $(2)$ gives,

$\begin{aligned} \dfrac{47 - 4b}{3} &=& a+b \\ \implies 3a + 7b &=& 47 \end{aligned}$

Which gives,

$a = \boxed{11} \\ b = \boxed{2}$

as the only prime solution.

Therefore,

$c = \boxed{7}$