How do you solve #x^2 - 6x + 4 = 0# by completing the square ?

1 Answer

Answer:

Add a constant to both sides such that the left hand side is a perfect square. #x = 3+sqrt5 = 3+-(2.236)#

Explanation:

In order to complete the square, we must determine what on the left hand side could be squared to give us the given #x^2# and #x# coefficients. A squared binomial would take the form #(ax+b)^2 = a^2x^2 + 2abx + b^2#. Looking at our equation, our #a# is going to be 1, which means that #2b = -6 -> b = -3#.

Thus, our squared binomial will take the form #(x-3)^2 = x^2 - 6x + 9#

In order to get this polynomial on the left hand side, we add 5 to both sides .

#x^2-6x+4+5 = 0 +5 -> x^2-6x+9 = 5 rarr (x-3)^2=5#

From here, we can square root the left hand side.

#-> (x-3)^2 = 5 -> +-sqrt((x-3)^2) = +-sqrt(5) -> (x-3) =+- sqrt5#

Then adding 3 to both sides:

#x = 3+-sqrt5 approx 3+-2.236 approx 0.764 or 5.236#