Question

what will be 1/alpha-1/beta

Answer 1

The general quadratic equation is

[math]\quad ax^2+bx+c\ =\ 0.[/math]

Because the equation is quadratic, we know that [math]\ a\ne 0\ [/math] so we can divide through by [math]a[/math] to give an equivalent form:

[math]\quad x^2+\frac{b}{a}x+\frac{c}{a}\ =\ 0[/math]

and this equation has roots [math]\alpha[/math] and [math]\beta[/math] which means that

[math]\begin{align}\quad x^2+\tfrac{b}{a}x+\tfrac{c}{a}\ & =\ (x-\alpha)(x-\beta)\\ & =\ x^2-(\alpha+\beta)x+(\alpha\beta)\end{align}[/math]

and equating the coefficients yields Vieta’s formulae:

[math]\quad \alpha+\beta\ =\ \frac{-b}{a}\qquad[/math] and [math]\qquad \alpha\beta\ =\ \frac{c}{a}[/math]

Now we can answer the question:

[math]\quad \boxed{\tfrac{1}{\alpha}+\tfrac{1}{\beta}\ =\ \tfrac{\alpha+\beta}{\alpha\beta}\ =\ \tfrac{-b}{c}}[/math]

Interestingly, we also have [math]\ \frac{1}{\alpha\beta}=\frac{a}{c}\ [/math] which means that [math]\ \frac{1}{\alpha}\ [/math] and [math]\ \frac{1}{\beta}\ [/math] are the roots of

[math]\quad cx^2+bx+a\ =\ 0.[/math]

In words: if you reverse the coefficients of your quadratic then you get the reciprocals of the roots.