Question

Let $f\left(x\right)=x^2-px+q,$ p is an odd positive integer and the roots of the equation $f\left(x\right)=0$ are two distinct prime numbers, if $p+q=35$, then the value of $f\left(10\right)\left({\sum }_{r=1}^{10}f\right(r\left)\right)-878$ equal to ___

Answer 1

$\because$ Roots of $x^2-px+q=0$ are two distinct prime numbers.
Sum of roots = p = odd positive integer (given)
$\Rightarrow$ Both roots of the equation cannot be odd.
$\therefore$ One root must be 2(prime number)
Then $(2{)}^2-p(2)+q=0$
or $2p-q=4$
and given $p+q=35$
Solving Eqn. (i) and (ii), we get
$p=13,q=22$
$\therefore f\left(x\right)=x^2-13x+22$
$\therefore {\sum }_{r=1}^{10}{r}^2-13{\sum }_{r=1}^{10}r+22{\sum }_{r=1}^{10}1$
$=\frac{\mathrm{10.11.21}}{6}-\frac{\mathrm{13.10.11}}{2}+22.10$
$=-110$ and $f\left(10\right)=-8$
$\therefore f\left(10\right){\sum }_{r=1}^{10}f\left(r\right)=(-8)(-100)=880$