## Monday, August 17, 2015

### Evaluate a+2b.

Suppose that there exist two positive real numbers $a$ and $b$ such that $(a-2) (a^2+2a-18)+ab(2a+b+18)-96+2(b-2)(b+3)(b+2)=0$.

Evaluate $a+2b$.

My solution:

We all like to see when the given equation is already in the factored form, because we learned how to factorize the polynomials since we were in younger grade, we were taught in order to solve for equation, we need to factorize it so we would be one step closer in achieving our goal, that is to solve for the roots of the given equation.

Albeit the given problem looks like it is in a pretty good partially factorized format, it doesn't help us at all to find the value for the intended expression of $a+2b$.

Why? Simply because there is nothing much left to do of that equation.

Instead, we need to expand the given partially factored form of the equation and see if we could factor it the right way, so we could get one of the factor carries the sum of $a$ and $2b$.

Upon expanding of the given equation, we see that we have:

$a^3+2 a^2 b+a b^2+18 a b-22 a+2 b^3+6 b^2-8 b-84=0$

We know we must "force" the term $a+2b$ out of it, in fact, we could still exploit what we're given, to let $a=2$ and $b=2$ and substitute them into the original equation and see what we have gotten:

$(a-2) (a^2+2a-18)+ab(2a+b+18)-96+2(b-2)(b+3)(b+2)$

$=(2-2) (a^2+2a-18)+2(2)(2(2)+2+18)-96+2(2-2)(b+3)(b+2)$

$=0+4(24)-96+0$

$=0$

Bingo! That means $a=b=2$ is one of the solution of the equation $(a-2) (a^2+2a-18)+ab(2a+b+18)-96+2(b-2)(b+3)(b+2)=0$ and so $a+2b=2+2(2)=6$, and we're done!

But hey, wait a minute, what if the problem has more than one solution, isn't we need to find them all out? Or if the problem has only one such solution, we should and need to prove that is the case!

Yes, you have to prove either case is right, then you can be said you have solved the problem completely.

The one surest way to proceed is to perform the polynomial long division to the expression $a^3+2 a^2 b+a b^2+18 a b-22 a+2 b^3+6 b^2-8 b-84$ by $a-2b-6$, and we see that we obtain:

$(a-2) (a^2+2a-18)+ab(2a+b+18)-96+2(b-2)(b+3)(b+2)=0$

$a^3+2 a^2 b+a b^2+18 a b-22 a+2 b^3+6 b^2-8 b-84=0$

$(a+2b-6) (a^2+6a+b^2+6b+14)=0$

$(a+2b-6) ((a+3)^2+(b+3)^2-18+14)=0$

$(a+2b-6) ((a+3)^2+(b+3)^2-4)=0$

Note that the second factor when we set $(a+3)^2+(b+3)^2-4=0$ is a circle centered at $(-3,\,-3)$ with radius of $2$. That means it is a circle lies in the third quadrant with $(a+3)^2+(b+3)^2-4\lt 0$, that implies the only way $(a+2b-6) ((a+3)^2+(b+3)^2-4)=0$ can be zero iff $a+2b-6=0$, i.e. $a+2b=6$, and we're now done.