JEE Main 2024 (Online) 1st February Evening ShiftQuestion 1
Let $\alpha$ and $\beta$ be the roots of the equation $px^{2}+qx-r=0$, where $p\neq0$. 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 :
Quadratic Equation · Relation Between Roots and Coefficients
Solution
Concept. With $q=pk$, $r=pk^2$ the equation becomes $x^2+kx-k^2=0$. Formulas used. (1) $\frac1\alpha+\frac1\beta=\dfrac{\alpha+\beta}{\alpha\beta}$ and $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$Step 1 - Sum and product. $\alpha+\beta=-k$, $\alpha\beta=-k^2$. Step 2 - Find $k$. By (1), $\frac{-k}{-k^2}=\frac1k=\frac34$, so $k=\frac43$. Step 3 - Value. $(\alpha-\beta)^2=k^2+4k^2=5k^2=\frac{80}{9}$. Answer: (B)