Question 68138
I had a problem I was working on with the quadratic equation rule.
This is how you solve it by completing the square.
{{{x^2+4x+2=0}}}
{{{x^2+4x+2-2=0-2}}}
{{{x^2+4x=-2}}}
{{{x^2+4x+4=4-2}}}
{{{x^2+4x+4=2}}}
{{{(x+2)^2=2}}}
{{{sqrt((x+2)^2)=+-sqrt(2)}}}
{{{x+2=+-sqrt(2)}}}
{{{x+2-2=-2+-sqrt(2)}}}
{{{highlight(x=-2+-sqrt(2))}}}
:
If you were supposed to solve it using the quadratic formula:
{{{x=(-b+-sqrt(b^2-4ac))/2a}}}
a=1, b=4 and c=2
{{{x=(-4+-sqrt((4)^2-4(1)(2)))/(2(1))}}}
{{{x=(-4+-sqrt(16-8))/2}}}
{{{x=(-4+-sqrt(8))/2}}}
{{{x=(-4+-sqrt(4)*sqrt(2))/2}}}
{{{x=(-4+or-2*sqrt(2))/2}}}
{{{x=-4/2+or-2*sqrt(2)/2}}}
{{{highlight(x=-2+-sqrt(2))}}}
Either way you eventually get there.
Happy Calculating!!!!