Question 7474
ok, completing the square...


{{{x^2+6x-7}}}. First ignore the 7. Just think about the {{{x^2+6x}}}. What i want to do is think of the correct number (not the 7) that needs to be there so that I can factorise the quadratic as {{{(x+k)^2}}} where k is a "nice" number, usually an integer, sometimes a simple fraction, rarely a horrible number.


think about {{{(x+k)^2}}}. This expands to be {{{x^2 + 2kx + k^2}}}, so we look at the coefficient before the x-term, our +6, and we note that this is like the "2k", so k must be 3. The number we therefore require {{{(k^2)}}} is +9, so what we need is:


{{{x^2+6x+9}}}.


That is it done, really. All we have to do now is get this to look like the original, {{{x^2+6x-7}}}, as follows:


{{{x^2+6x+9-9-7}}}...note the +9-9 cancel each other out, leaving the -7
{{{x^2+6x+9 - 16}}}
{{{(x+3)^2 - 16}}}


Hope this is clear?


And now to answer the question... {{{(x+3)^2 - 16 = 0}}}


{{{(x+3)^2 = 16}}}
{{{(x+3) = sqrt(16)}}} or {{{(x+3) = - sqrt(16)}}}


so x+3 = 4 OR x+3 = -4
so x=1 OR x=-7


jon.