# 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!

Sep 18, 2017

$m + n = - b$

#### Explanation:

Suppose we have a quadratic function $f \left(x\right) = a {x}^{2} + b x + c$.

If $m$ and $n$ are the $x$-intercepts of $f \left(x\right)$, then $\left(x - m\right) \left(x - n\right) = 0$.

Let's expand the parentheses to get:

$R i g h t a r r o w \left(x - m\right) \left(x - n\right) = 0$

$R i g h t a r r o w \left(x\right) \left(x\right) + \left(x\right) \left(- n\right) + \left(- m\right) \left(x\right) + \left(- m\right) \left(- n\right) = 0$

$R i g h t a r r o w {x}^{2} - n x - m x + m n = 0$

$R i g h t a r r o w {x}^{2} - \left(m + n\right) x + m n = 0$

This is now in the form of a quadratic equation.

Let's compare this to $f \left(x\right)$.

If $a = 1$, we get $f \left(x\right) = {x}^{2} + b x + c$.

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

$R i g h t a r r o w - \left(m + n\right) = b$

$\mathmr{and}$

$R i g h t a r r o w m n = c$

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

$R i g h t a r r o w - \left(m + n\right) = b$

$\therefore m + n = - b$

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 $a = 1$.