Question

If the roots of the equation $x^3+P{x}^2+Qx-19=0$ are each one more than the roots of the equation $x^3-A{x}^2+Bx-C=0$, where $A,B,C,P$ and $Q$ are constants then the value of $A+B+C=$

• A. $18$
• B. $20$
• C. $22$
• D. $16$

Answer 1

Answer: (A)

$18$

$\Rightarrow$ Let the roots of $x^3-A{x}^2+Bx-C=0$ be $\alpha ,\beta ,\gamma$

$\Rightarrow$ Now, $\alpha +\beta +\gamma =\frac{-Coeff{x}^2}{Coeff{x}^3}=\frac{-(-A)}{1}=A$ ---- ( 1 )

$\Rightarrow$ $\alpha \beta +\beta \gamma +\alpha \gamma =\frac{Coeff.ofx}{Coeff.x^3}=B$ ------- ( 2 )

$\Rightarrow$ $\alpha \beta \gamma =\frac{Coeff.1}{Coeff.x^3}=C$ -------- ( 3 )

$\Rightarrow$ According to the question, roots of the equation $x^3+p{x}^2+q-19=0$ are $(\alpha +1),(\beta +1),(\gamma +1)$

$\Rightarrow$ As in the above example $-p=\alpha +1+\beta +1+\gamma +1=\alpha +\beta +\gamma +3$

$19=(\alpha +1)(\beta +1)(\gamma +1)$ [ Product of roots ]

$19=\alpha \beta \gamma +\alpha \beta +\beta \gamma +\alpha \gamma +\alpha +\beta +\gamma +!$

$19=A+B+C+1$ [ From ( 1 ), ( 2 ) and ( 3 ) ]

$\therefore$ $A+B+C=18$