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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?

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    $\begingroup$ 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. $\endgroup$
    – Taladris
    Aug 23, 2021 at 2:49
  • $\begingroup$ Put another way, the discriminant is a square, $9.$ That means it factors nicely, integer coefficients $\endgroup$
    – Will Jagy
    Aug 23, 2021 at 2:51

1 Answer 1

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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.

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