We are able to solve this problem by expanding, however, I've seen a shortcut method but I'm not sure how this is done. (perhaps comparing coefficients)
The shortcut method used is:
$(15+24)^2=12∗51−24^2$ <- how does this work?
$(15+24)^2=36$
$(15y+24)$ = $+6$ or $-6$
$y$ = $-2$ or $-1.2$ which gives a correct answer
Anyone could explain the shortcut method used? Why subtract $24^2$ from $12*51$? This method completely ignores the terms $20y^2$ and $64y$, but it works for many other quadratics, so I believe there must be some sort of reasoning behind this.
(I'm not looking for answers to solve this problem but possibly an explanation for the method stated)
Although, this is not a well posed question and there absolutly nothing fancy about this so called "method", I will nevertheless answer it just to show you that in math nothing comes for free. If you have a "method" where two coefficients can be dropped there have to be heavy assumptions! Lets get to it: For $A\neq0$ we have $$ (Ay+B)^2=Cy^2+Dy+E\\ =\frac{C}{A^2}(Ay+B)^2- \frac{2CB}{A}y-\frac{CB^2}{A^2}+ Dy+ E $$ Now if $D= \frac{2CB}{A}$ and ${A^2\neq C}$ this simplifies to $$ (Ay+B)^2=\frac{C}{A^2}(Ay+B)^2-\frac{CB^2}{A^2}+ E \\ \Leftrightarrow (1-\frac{C}{A^2})(Ay+B)^2=-\frac{CB^2}{A^2}+ E\\ \Leftrightarrow (A^2-C)(Ay+B)^2=-CB^2+ EA^2\\ \Leftrightarrow (Ay+B)^2=\frac{EA^2-CB^2}{A^2-C}\\ $$ Now if we further assume that $E=B^2 (2-\tfrac{A^2}{C})$ we miraculously (yes, that's sarcasm) end up with $$ (Ay+B)^2=E-B^2. $$
So to summarize, this "methods" works only if $$ A\neq0\\ D= \frac{2CB}{A}\\ {A^2\neq C}\\ E=B^2 (2-\tfrac{A^2}{C}) $$