How do you solve # a^2 - 4a - 21 = 0# by completing the square?

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Answer 1 · 2016-05-24T15:25:59

The solutions are:
#color(green)(a = 7# , # color(green)(a = -3#

#a^2 -4a - 21 = 0#

#a^2 -4a = 21#

To write the Left Hand Side as a Perfect Square, we add 4 to both sides
#a^2 -4a + color(green)(4) = 21 + color(green)(4#

#a^2 - 2 * a * 2 + 2^2 = 25#

Using the Identity #color(blue)((a-b)^2 = a^2 - 2ab + b^2#, we get

#(a-2)^2 = 25#

#a-2 = sqrt25# or #a-2 = -sqrt25#

#a-2 = 5# or #a-2 = -5#

#a = 5+2# or #a = -5+2#

#color(green)(a = 7# , # color(green)(a = -3#