Question

$x^2+px+45=0$ has one root as ${}^{10}{C}_2$,then value of $p$ is

• A. $46$
• B. $45$
• C. $-45$
• D. $-46$

Answer 1

The correct option is D $-46$

$f\left(x\right)=x^2+px+45$ has root as ${}^{10}{c}_2$

${}^{10}{c}_2=\frac{10!}{(10-2)!2!}=\frac{10!}{8!2!}=\frac{10\ast 9\ast 8!}{8!\ast 2!}=45f\left(45\right)=0{45}^2+45p+45=045p=-45\ast 45-45p=\frac{-45(45+1)}{45}p=-46$