Question 708291
{{{abs(x-11) = x^2-11x}}}
If this were
{{{abs(x-11) = 35}}}
would it be easier? You probably know that this would be solved using two equations:
{{{x-11 = 35}}} or {{{x-11 = -35}}}<br>
Well, the problem we have can be solved in <i>exactly the same way!</i>
{{{x-11 = x^2-11x}}} or {{{x-11 = -(x^2-11x)}}}
The trick part is that we need the negative of the whole right side in the second equation. When the right side is just a number, like 35, you can just stick a "-" in front.  But with multiple terms, like {{{x^2-11x}}}, we need to put parentheses around the right side. With the parentheses, the "-" in front will negate the entire expression, not just the {{{x^2}}}.<br>
We now have two quadratic equations to solve. Let's do them one at a time. With the first equation we want one side to be zero. Subtracting the entire left side from both sides we get:
{{{0 = x^2-11x+11}}}
This will not factor we we will use the Quadratic Formula:
{{{x = (-(-11) +- sqrt((-11)^2-4(1)(11)))/2(1)}}}
Simplifying...
{{{x = (-(-11) +- sqrt(121-4(1)(11)))/2(1)}}}
{{{x = (-(-11) +- sqrt(121-44))/2(1)}}}
{{{x = (-(-11) +- sqrt(77))/2(1)}}}
{{{x = (11 +- sqrt(77))/2}}}
which is short for:
{{{x = (11 + sqrt(77))/2}}} or {{{x = (11 - sqrt(77))/2}}}<br>
Now for the second equation we got earlier:
{{{x-11 = -(x^2-11x)}}}
Before we get the zero, let's simplify the right side:
{{{x-11 = -x^2+11x)}}}
Adding {{{x^2}}} and subtracting 11x we get:
{{{x^2-11x-11 = 0}}}
Again this will not factor so again we use the formula:
{{{x = (-(-11) +- sqrt((-11)^2-4(1)(-11)))/2(1)}}}
Simplifying...
{{{x = (-(-11) +- sqrt(121-4(1)(-11)))/2(1)}}}
{{{x = (-(-11) +- sqrt(121+44))/2(1)}}}
{{{x = (-(-11) +- sqrt(165))/2(1)}}}
{{{x = (11 +- sqrt(165))/2}}}
which is short for:
{{{x = (11 + sqrt(165))/2}}} or {{{x = (11 - sqrt(165))/2}}}<br>
So there are four solutions to your equation:
{{{x = (11 + sqrt(77))/2}}} or {{{x = (11 - sqrt(77))/2}}}
{{{x = (11 + sqrt(165))/2}}} or {{{x = (11 - sqrt(165))/2}}}