Question

If $\alpha ,\beta$ and $\gamma$ are the zeroes of a polynomial $x^3-5x^2-2x+24,$ such that $\alpha \beta =12$ then difference of first two zeroes can be

• A. $1$
• B. $-1$
• C. $0$
• D. $-2$

Answer 1

Answer: (A), (B)

$-1$
$1$

$f\left(x\right)=x^3-5x^2-2x+24$

Let $\alpha \beta =12$

$\alpha +\beta +\gamma =-(-5)=5$

$\alpha \beta +\beta \gamma +\alpha \gamma =\frac{-2}{1}=-2$

$\alpha \beta \gamma =-\frac{24}{1}=-24$

$\alpha \beta =12⟹12\gamma =-24=\gamma =-2$

$\alpha +\beta +\gamma =5$

$\alpha +\beta -2=5⟹\alpha +\beta =7$

$(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta =49-4\times 12⟹\alpha -\beta =\pm 1$

$\alpha +\beta =7$ and $\alpha -\beta =1$

(or) $\alpha +\beta =7$ and $\alpha -\beta =1$

On solving we get:

$\alpha =4,\beta =3$ (or) $\alpha =3,\beta =4$

The difference between the roots is $4-3=1$ or $3-4=-1$