As taught at GCSE, every single quadratic can be expressed in this form:

$$ax^2 + bx + c = 0$$

If we divide throughout by $ a $, we can derive:

$$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$

If we assign $ \alpha $ to the value of one root and $ \beta $ to the other, we can derive that:

$$(x - \alpha)(x - \beta) = 0$$

By expanding this quadratic, we obtain:

$$x^2 - \alpha x - \beta x + \alpha\beta = 0$$

Which simplifies to:

$$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$

From this, we can derive the sum and product of the roots from the first equation:

Sum of roots

$\alpha + \beta = \frac{-b}{a}$

Product of roots

$\alpha \times \beta = \frac{c}{a}$

New equations by substitution

If new roots to be found are reciprocals of the original equation’s roots, then the new equation can be derived by making the substitutions:

$$x \rightarrow \frac{1}{y}$$ $$y \rightarrow \frac{1}{x}$$

Useful derivations

These can be easily derived, but they are easier to remember:

$$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \times \beta}$$ $$\alpha^2\beta^2 = (\alpha\beta)^2$$ $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$$ $$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$$