What is the quadratic formula for #f(b)=b^2 - 4b + 4 = 0#?

1 Answer
Oct 22, 2015

Answer:

Rewriting #f(b)# as #f(x)# will permit you to use the standard formula with less confusion (since the standard quadratic formula uses #b# as one of its constants)

Explanation:

(since the given equation uses #b# as a variable, we will need to express the quadratic formula, which normally uses #b# as a constant, with some variant, #hatb#.

To help reduce confusion, I will rewrite the given #f(b)#as
#color(white)("XX")f(x)=x^2-4x+4=0#

For the general quadratic form:
#color(white)("XX")hatax^2+hatbx+hatc=0#
the solution given by the quadratic equation is
#color(white)("XX")x=(-hatb+-sqrt(hatb^2-4hatahatc))/(2hata)#

With #hata = 1#, #hatb=-4#, and #hatc=+4#
we get
#color(white)("XX")b=(x=)(4+-sqrt((-4)^2+4(1)(4)))/(2(1))#
as the quadratic formula