Question

If \(\alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then \[ \left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2 = \]

• A. \(\dfrac{b^2(b^2-4ac)}{a^2c^2}\)
• B. \(\dfrac{b^2(b^2-4ac)}{a^4}\)
• C. \(\dfrac{b^2(b^2-4ac)}{ca^3}\)
• D. \(\dfrac{b^2(b^2-4ac)}{c^4}\)

Hint:

Useful relations: \[ \boxed{ \alpha+\beta=-\frac{b}{a},\qquad \alpha\beta=\frac{c}{a} } \]

Answer 1

The Correct Option is A

Step 1: Note the two facts we get from the equation.
For $ax^2+bx+c=0$ with roots $\alpha,\beta$, we always have $\alpha+\beta=-\frac{b}{a}$ and $\alpha\beta=\frac{c}{a}$. We want $\left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2$.
Step 2: Combine the fraction before squaring.
$\frac{\alpha}{\beta}-\frac{\beta}{\alpha}=\frac{\alpha^2-\beta^2}{\alpha\beta}=\frac{(\alpha-\beta)(\alpha+\beta)}{\alpha\beta}$, so the square is $\dfrac{(\alpha-\beta)^2(\alpha+\beta)^2}{(\alpha\beta)^2}$.
Step 3: Substitute the sum and product.
Using $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\frac{b^2}{a^2}-\frac{4c}{a}=\frac{b^2-4ac}{a^2}$, and $(\alpha+\beta)^2=\frac{b^2}{a^2}$, $(\alpha\beta)^2=\frac{c^2}{a^2}$, multiply and divide to get $\dfrac{b^2-4ac}{a^2}\cdot\dfrac{b^2}{a^2}\Big/\dfrac{c^2}{a^2}=\dfrac{b^2(b^2-4ac)}{a^2c^2}$.
\[ \boxed{\dfrac{b^2(b^2-4ac)}{a^2c^2}} \]