ex1.5_11

$\quad x^2+4x+3=0$

$a=1$ $b=4$ $c=3$ . It follows that

h = -\frac{4}{2} = -2, \mbox{ and}\\[0.25in] k = (-2)^2+4(-2)+3 = -1.

Therefore, the completed square is

x^2+4x+3 = (x+2)^2 -1.

Now we solve by isolating the square and taking square roots:

(x+2)^2-1=0,\\[0.25 in] (x+2)^2 = 1, \\[0.25 in] x+2 = \pm 1, \\[0.25 in] x+2 = -1 \quad \mbox{or} \quad x+2 = 1,\\[0.25 in] x=-3 \quad \mbox{or} \quad x=-1.