Question

In an equation `ax^2+bx+c`, where `(m,n)` are the x-intercepts, what is `(m,n)` in terms of `b`?

A detailed explanation of how you get there would be very much appreciated!

Answer 1

`m + n = - b`

Explanation:

Suppose we have a quadratic function `f(x) = a x^(2) + b x + c`.

If `m` and `n` are the `x`-intercepts of `f(x)`, then `(x - m)(x - n) = 0`.

Let's expand the parentheses to get:

`Rightarrow (x - m)(x - n) = 0`

`Rightarrow (x)(x) + (x)(- n) + (- m)(x) + (- m)(- n) = 0`

`Rightarrow x^(2) - n x - m x + m n = 0`

`Rightarrow x^(2) - (m + n) x + m n = 0`

This is now in the form of a quadratic equation.

Let's compare this to `f(x)`.

If `a = 1`, we get `f(x) = x^(2) + b x + c`.

Comparing coefficients with our quadratic equation, we can deduce that:

`Rightarrow - (m + n) = b`

`and`

`Rightarrow m n = c`

We need to express `(m, n)` in terms of `b`, so let's only consider the first equation:

`Rightarrow - (m + n) = b`

`therefore m + n = - b`

This comes from Vieta's theorem of quadratic polynomials.

It states that the sum of the roots `x_(1)` and `x_(2)` of a quadratic polynomial `a x^(2) + b x + c` are equal to `- b`, if `a = 1`.