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!

1 Answer
Sep 18, 2017

#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#.