Question

If $1,2,3$ and $4$ are the roots of the equation $x^4+a{x}^3+b{x}^2+cx+d=0$, then $a+2b+c=$

• A. $-25$
• B. $0$
• C. $10$
• D. $24$

Answer 1

The correct option is C $\left(10\right)$

As $1,2,3,4$ are roots of $x^4+a{x}^3+b{x}^2+cx+d=0$

Gives $S_1=$sum of roots=$-\frac{\text{coefficient of~}{x}^3}{\text{coefficient of}{x}^4}=1+2+3+4=-a$
$\Rightarrow a=-10$
Similar way, $S_2=1\cdot 2+1\cdot 3+1\cdot 4+2\cdot 3+2\cdot 4+3\cdot 4=b\Rightarrow b=35S_3=1\cdot 2\cdot 3+1\cdot 2\cdot 4+2\cdot 3\cdot 4+3\cdot 4\cdot 1=-c\Rightarrow c=-50$

Therefore, $a+2b+c=-10+2\left(35\right)-50=-10+70-50=10$