Question

If two roots of the equation $(x-1)(2x^2-3x+4)=0$ coincide with roots of the equation $x^3+(a+1)x^2+(a+b)x+b=0$, where $a,b\in R$, then $2(a+b)$ is equal to

• A. $4$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

$(x-1)(2x^2-3x+4)=0$ can be written as $2x^3-5x^2+7x-4=0$
Let $\alpha ,\beta ,1$ be the roots of the equation $(x-1)(2x^2-3x+4)=0$
Then $\alpha +\beta +1=-\frac{-5}{2}$
$\Rightarrow \alpha +\beta =\frac{3}{2}$
Also, $\alpha \beta =2$
Now, let $\alpha ,\beta ,\gamma$ be the roots of the equation $x^3+(a+1)x^2+(a+b)x+b=0$
Then $\alpha \beta \gamma =-b$
$\Rightarrow \gamma =\frac{-b}{2}$
Also, $\alpha +\beta +\gamma =-(a+1)$
$\Rightarrow b=2a+5$ ......(i)
Also, $\alpha \beta +\beta \gamma +\gamma \alpha =a+b$
$\Rightarrow 7b+4a=8$ ......(ii)
Solving (i) and (ii), we get
$\Rightarrow a=\frac{-3}{2},b=2$
$\Rightarrow 2(a+b)=1$