Question

explain the steps also
​

Answer 1

We know,

$\text{In the equation\ ax^2+bx+c=0, if \ x\in\{\alpha,\ \beta \}, }\\\\\text{then \ \alpha+\beta=-\dfrac{b}{a}\quad\&\quad\alpha\beta=\dfrac{c}{a}. }$

So,

$a=3\quad;\quad b=-5\quad;\quad c=4\\\\\\\alpha+\beta=-\dfrac{-5}{3}=\dfrac{5}{3}\\\\\\\alpha\beta=\dfrac{4}{3}$

Now,

$\begin{aligned}&\dfrac{1}{\alpha}+\dfrac{1}{\beta}-2\alpha\beta\\\\\implies\ \ &\dfrac{\alpha+\beta}{\alpha\beta}-2\alpha\beta\\\\\implies\ \ &\dfrac{\left(\dfrac{5}{3}\right)}{\left(\dfrac{4}{3}\right)}-2\cdot\dfrac{4}{3}\\\\\implies\ \ &\dfrac{5}{4}-\dfrac{8}{3}\\\\\implies\ \ &\mathbf{-\dfrac{17}{12}}\end{aligned}$