Question

given that alpha and beta are roots of the quadratic equation ax²+bx+c=0, show that alpha+beta=-6÷a and alphabeta=c÷a​

Answer 1

Answer:

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

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

Step-by-step explanation:

Given

$ax^2 + bx + c = 0$

$Roots: \alpha \& \beta$

Required

Show that:

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

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

$ax^2 + bx + c = 0$

Divide through by a

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

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

The general form of a quadratic equation is:

$x^2 - (Sum)x + (Product) = 0$

By comparison, we have:

$-(Sum)x = \frac{b}{a}x$

$-(Sum) = \frac{b}{a}$

Sum is calculated as:

$Sum = \alpha + \beta$

So, we have:

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

Divide both sides by -1

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

Similarly;

$Product = \frac{c}{a}$

Product is calculated as:

$Product = \alpha * \beta$

So, we have:

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

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