## anonymous 5 years ago Solve x2 + 4x – 12 = 0 by completing the square. Please help!

1. myininaya

$x^2+4x=12$ added 12 on both sides

2. myininaya

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

3. myininaya

$(x+2)^2=16$

4. myininaya

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

5. myininaya

$x+2 = \pm \sqrt{16}$ $x+2= \pm 4$

6. myininaya

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

7. myininaya

$x=2 \text{ or } x=-6$

8. anonymous

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

9. 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

10. anonymous

thanks guys! (: