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Mathematics 58 Online
OpenStudy (anonymous):

Solve x2 + 4x – 12 = 0 by completing the square. Please help!

myininaya (myininaya):

\[x^2+4x=12\] added 12 on both sides

myininaya (myininaya):

\[x^2+4x+4=12+4\] added 4 on both sides to complete the square on the left hand side

myininaya (myininaya):

\[(x+2)^2=16\]

myininaya (myininaya):

Now take square root of both sides don't forget the plus or minus

myininaya (myininaya):

\[x+2 = \pm \sqrt{16}\] \[x+2= \pm 4\]

myininaya (myininaya):

Now subtract 2 on both sides \[x=-2 \pm 4\] this means we have two values for x \[x=-2+4 \text{ or } x=-2-4\]

myininaya (myininaya):

\[x=2 \text{ or } x=-6\]

OpenStudy (anonymous):

\[x ^{2}+4x+4-16=0 \rightarrow (x+2)^2=16 \rightarrow x+2= \pm 4;\]; x=2 and x=-6

hero (hero):

x2 + 4x – 12 = 0 x^2+6x-2x-12 =0 x(x+6)-2(x+6) = 0 (x+6)(x-2) = 0 x = -6,2

OpenStudy (anonymous):

thanks guys! (:

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