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

`\alpha + \beta = -(b)/(a)`

`\alpha \beta = (c)/(a)`

Explanation:

Given

`ax^2 + bx + c = 0`

`Roots: \alpha \& \beta`

Required

Show that:

`\alpha + \beta = -(b)/(a)`

`\alpha \beta = (c)/(a)`

`ax^2 + bx + c = 0`

Divide through by a

`(a)/(a)x^2 + (b)/(a)x + (c)/(a) = (0)/(a)`

`x^2 + (b)/(a)x + (c)/(a) = 0`

The general form of a quadratic equation is:

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

By comparison, we have:

`-(Sum)x = (b)/(a)x`

`-(Sum) = (b)/(a)`

Sum is calculated as:

`Sum = \alpha + \beta`

So, we have:

`-(\alpha + \beta) = (b)/(a)`

Divide both sides by -1

`\alpha + \beta = -(b)/(a)`

Similarly;

`Product = (c)/(a)`

Product is calculated as:

`Product = \alpha * \beta`

So, we have:

`\alpha * \beta = (c)/(a)`

`\alpha \beta = (c)/(a)`