Question 19274
right, fairly easy concept here. As follows:


{{{x^2-6x=16}}} Just look at the left hand side: {{{x^2-6x}}}. We need to figure out which number to add/subtract such that we can then factorise it as {{{x+a)^2}}} or {{{(x-a)^2}}}


so, {{{x+a)^2}}} expanded would give {{{x^2 + 2ax + a^2}}} and the negative version would have -2ax instead.


So, we have -6x which is the (-2ax) which means that a needs to be 3. 


So, the {{{a^2}}} term is {{{3^2}}} ---> 9


So, we have to add 9...to both sides remember, to keep the equation unaltered... {{{x^2-6x+9 = 25}}}

So, {{{x^2 - 6x + 9}}} can be written instantly as {{{(x-3)^2}}} --> the negative version, since the x-term was negative.


and we have {{{(x-3)^2 = 25}}}
{{{(x-3) = +sqrt(25)}}} OR {{{(x-3) = -sqrt(25)}}}


--> x-3 = +5   OR   x-3 = -5
--> x = 8   OR   x = -2


jon.