How do you solve the equation #x^2+1.4x+0.49=0.81# by completing the square?

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Answer 1 · 2016-12-18T09:28:47

#x = 0.2" "# or #" "x = -1.6#

We will use the difference of squares identity, which can be written:

#a^2-b^2 = (a-b)(a+b)#

Given:

#x^2+1.4x+0.49 = 0.81#

Both the left hand and right hand side of this equation are perfect squares already:

#(x+0.7^2) = x^2+1.4x+0.49 = 0.81 = 0.9^2#

Hence:

#0 = (x+0.7^2)-0.9^2#

#color(white)(0) = ((x+0.7)-0.9)((x+0.7)+0.9)#

#color(white)(0) = (x-0.2)(x+1.6)#

Hence:

#x = 0.2" "# or #" "x = -1.6#