Question 72679
If (a) does not equal (b), solve for x in terms of (a) and (b):
{{{x^2+ax=-b(a+x)}}} get rid of parens

{{{x^2+ax=-ab-bx}}}  add ab and bx to both sides

{{{x^2+ax+bx+ab=0}}}  rearrange into standard form for a quadratic eq.

{{{x^2+(a+b)x+ab=0}}}  This is a quadratic equation in standard form where"
{{{A=1}}}
{{{B=(a+b)}}}
{{{C=ab}}}

By inspection, we can see that the equation can be factored because the B coefficient is the sum of the product of the C coefficient.  This is always the case, when A=1.  So we have:

{{{(x+a)(x+b)=0}}}

{{{x+a=0}}}
{{{x=-a}}}
and
{{{x+b=0}}}
{{{x=-b}}}


  CK
{{{x^2+ax=-b(a+x)}}}
{{{a^2-a^2=-ab+ab}}}
{{{0=0}}}
and
{{{b^2-ab=-ab+b^2}}} or
{{{b^2-ab=b^2-ab}}}



Hope this helps-------------------ptaylor