# Is there a method to factor equations with two variables raised to the second power? [closed]

I found the equation $$2b^2-ab-a^2=0$$ on a problem and couldn't find a way to factor it. Is there any method to factor these types of equations?

• You can consider that $a$ is fixed and that the expression $2b^2-ab-a^2=p(b)$ as a polynomial in $b$. $p$ has an obvious zero, so you can factor it. Aug 23, 2021 at 2:49
• Put another way, the discriminant is a square, $9.$ That means it factors nicely, integer coefficients Aug 23, 2021 at 2:51

An easy way to do it is by defining a variable $$u=\frac{a}{b}$$ and then dividing by $$b^2$$ on both sides (assuming $$b\neq 0$$). We will then get a quadratic in $$u$$ e.g. $$2b^2-ab-a^2=0$$ $$2-\frac{a}{b}-\frac{a^2}{b^2}=0$$ $$2-u-u^2=0$$ $$u^2+u-2=0$$ $$(u+2)(u-1)=0$$ $$u=1,-2$$ $$a=b,-2b$$ We also have the potentially singular solution that results when $$b=0$$. We can easily see that the only solution of that form is $$a,b=0$$, which is included in our general solution.
Another method to solve these types of equations where all terms are order $$2$$ is to consider one variable as a "constant" and solve using quadratic formula (or if you are bold, you can try for a factorization). For example, if we treat $$b$$ as a constant in the example you provided, $$a^2+ba-2b^2=0$$ $$a=\frac{-b\pm\sqrt{b^2+8b^2}}{2}$$ $$a=\frac{-b\pm 3b}{2}$$ $$a=b,-2b$$ This will be pretty similar in terms of difficulty as the former method. This method also has applications when solving equations in two variables where the equation is, say, a quartic in one variable and a quadratic in the other. You can treat the quartic variable as a constant and solve using quadratic equation. I can't find any examples at the moment, but I'll add them later.