Question

lf $1,1,\alpha$ are the roots of $x^3-6x^2+9x-4=0$, then $\alpha =$

• A. $-4$
• B. $4$
• C. $-3$
• D. $3$

Answer 1

Answer: (B)

$4$

Consider a cubic equation $a{x}^3+b{x}^2+cx+d=0$

Hence, if the roots are $\alpha ,\beta ,\gamma$, then

$\alpha +\beta +\gamma =\frac{-b}{a}$
$\Rightarrow \alpha .\beta +\alpha .\gamma +\beta .\gamma =\frac{c}{a}$

$\Rightarrow \alpha \beta \gamma -\frac{-d}{a}$

Now it is given that the roots are $1,1,\alpha$.

Hence, $1\times 1\times \alpha =\frac{-d}{a}$

$\Rightarrow \alpha =\frac{-d}{a}=\frac{-(-4)}{1}$

$\Rightarrow \alpha =4$