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

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Answer 1 · 2017-09-18T11:24:45

$m + n = - b$

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 $a = 1$ .