Question

If $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$, then the value ${(\frac{\alpha }{\beta }-\frac{\beta }{\alpha })}^2$ is

• A. $\frac{b^2(b^2-4ac)}{c^2{a}^2}$
• B. $\frac{b^2(b^2+4ac)}{c^2{a}^2}$
• C. $\frac{b^2(b^2-2ac)}{c^2{a}^2}$
• D. $\frac{b^2(b^2+2ac)}{c^2{a}^2}$

Answer 1

The correct option is A $\frac{b^2(b^2-4ac)}{c^2{a}^2}$

As $\alpha ,\beta$ are roots of $a{x}^2+bx+c=0$
then $\alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$

$\alpha -\beta =\sqrt{{(\alpha +\beta )}^2-4\alpha \beta }=\sqrt{\frac{b^2}{a^2}-\frac{4c}{a}}$

Now ${(\frac{\alpha }{\beta }-\frac{\beta }{\alpha })}^2={\left(\frac{{\alpha }^2-{\beta }^2}{\alpha \beta }\right)}^2=\frac{{(\alpha -\beta )}^2{(\alpha +\beta )}^2}{{\left(\alpha \beta \right)}^2}$

$=\frac{(\frac{b^2}{a^2}-\frac{4c}{a})\left(\frac{b^2}{a^2}\right)}{\left(\frac{c^2}{a^2}\right)}=\frac{(b^2-4ac)b^2}{a^2{c}^2}$