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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)

MPP
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1 Answers1

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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 $$ A\neq0\\ D= \frac{2CB}{A}\\ {A^2\neq C}\\ E=B^2 (2-\tfrac{A^2}{C}) $$

maxmilgram
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