Quadratic Equation Question 2
Question 2 - 2024 (01 Feb Shift 2)
Let $\alpha$ and $\beta$ be the roots of the equation $px^{2}+qx-r=0$, where $p \neq 0$. If $p, q$ and $r$ be the consecutive terms of a non-constant G.P and $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}$, then the value of $(\alpha-\beta)^{2}$ is :
(1) $\frac{80}{9}$
(2) 9
(3) $\frac{20}{3}$
(4) 8
Show Answer
Answer (1)
Solution
$px^{2}+qx-r=0« _{\beta}^{\alpha}$
$p=A, q=A R, r=A R^{2}$
$A x^{2}+A R x-A R^{2}=0$
$x^{2}+R x-R^{2}=0<{ } _{\beta}^{\alpha}$
$\because \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}$
$\therefore \frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4} \Rightarrow \frac{-R}{-R^{2}}=\frac{3}{4} \Rightarrow R=\frac{4}{3}$
$(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta=R^{2}-4\left(-R^{2}\right)=5\left(\frac{16}{9}\right)$
$=80 / 9$
