How do you solve #x^2 - 3x - 28 = 0# by completing the square?

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Answer 1 · 2016-06-19T05:40:29

x is either 7 or #-#4

We have, #x^2 -3x-28=0#

This can be written as #x^2 -3x = 28#

#x^2 - (2)(3/2)x = 28# (the trick is to take out 2 from the coefficient of x)

Now adding the square of 3/2 on both sides.

#x^2 - (2)(3/2)x + (3/2)^2 = 28 + (3/2)^2#

Carefully notice that the LHS becomes a square

#(x-3/2)^2 = 28 + 9/4#

[A side note: if #(f(x))^2 = g(x)# then #f(x) = +-sqrt(g(x))# ]

So, #x-3/2 = +-sqrt(28 + 9/4)#

#x-3/2 = +-sqrt(121/4)#

#x-3/2 = +-11/2#

#x= 3/2 +-11/2#

So x is either 7 or #-#4