How do you solve #x^2 - 6x + 9 = 0# using the quadratic formula?

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Answer 1 · 2016-07-10T21:38:50

x = 3

#x^2 - 6x + 9 = (x - 3)^2#
There ia a double-root at x = 3

Tony B

Using the formula as instructed.

Compare to: #ax^2+bx+c -> x=(-b+-sqrt(b^2-4ac))/(2a)#

In this case: #a=1; b=-6 and c=9# giving:

#x=(+6+-sqrt((-6)^2-4(1)(9)))/(2(1))#

#x=(6+-sqrt(36-36))/2#

#x=3+-0#

#x=3#