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

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Answer 1 · 2015-05-16T13:58:51

It's a rather strange way to solve it as it is easy to see that #x=0# and #x=-1# are both solutions. However, here's how you would do it...

#0 = 2x^2+2x = 2(x+1/2)^2 - 1/2#

Add 1/2 to both sides to get

#2(x+1/2)^2 = 1/2#

Divide both sides by 2 to get

#(x+1/2)^2 = 1/4#

So

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

Subtract 1/2 from both sides to get

#x=-1/2+-1/2=(-1+-1)/2#

Answer 2 · 2015-05-17T04:03:17

If #2x^2 + 2x = 0#, then #x^2 + x = 0#.

To complete this square, I need to square half of the coefficient of the 'x', which is 1/2.

#(1/2)^2=1/4# , so #x^2 + x + 1/4 = 1/4#

#(x + 1/2)^2=1/4#

#|x + 1/2| = sqrt(1/4)# (square roots can be positive or negative)

#|x + 1/2| = 1/2#

#x = {-1, 0}#