Unsure of reasoning behind a "shortcut" method to solve(15y+24)2=12(20y2+64y+51)

If you have a "method" where two coefficients can be dropped there have to be heavy assumptions! Lets get to it: For&nbsp;A≠0&nbsp;we have(Ay+B)2=Cy2+Dy+E=CA2(Ay+B)2−2CBAy−CB2A2+Dy+ENow if&nbsp;D=2CBA&nbsp;and&nbsp;A2≠C&nbsp;this simplifies to(Ay+B)2=CA2(Ay+B)2−CB2A2+E⇔(1−CA2)(Ay+B)2=−CB2A2+E⇔(A2−C)(Ay+B)2=−CB2+EA2⇔(Ay+B)2=EA2−CB2A2−CNow if we further assume that&nbsp;E=B2(2−tA2C)&nbsp;we miraculously (yes, that's sarcasm) end up with(Ay+B)2=E−B2.So to summarize, this "methods" works only ifA≠0D=2CBAA2≠C

"Unsure of reasoning behind a "shortcut" method to..." solved and explained

High School
Algebra
Algebra I
Quadratic function and equation

Kale Bright

Answered question

2022-04-09

Unsure of reasoning behind a "shortcut" method to solve

{(15 y + 24)}^{2} = 12 (20 y^{2} + 64 y + 51)

Answer & Explanation

szalbierzfytg

Beginner2022-04-10Added 13 answers

If you have a "method" where two coefficients can be dropped there have to be heavy assumptions! Lets get to it: For $A \neq 0$ we have
${(A y + B)}^{2} = C y^{2} + D y + E$

$= \frac{C}{A^{2}} {(A y + B)}^{2} - \frac{2 C B}{A} y - \frac{C B^{2}}{A^{2}} + D y + E$
Now if $D = \frac{2 C B}{A}$ and $A^{2} \neq C$ this simplifies to
${(A y + B)}^{2} = \frac{C}{A^{2}} {(A y + B)}^{2} - \frac{C B^{2}}{A^{2}} + E$

$\Leftrightarrow (1 - \frac{C}{A^{2}}) {(A y + B)}^{2} = - \frac{C B^{2}}{A^{2}} + E$

$\Leftrightarrow (A^{2} - C) {(A y + B)}^{2} = - C B^{2} + E A^{2}$

$\Leftrightarrow {(A y + B)}^{2} = \frac{E A^{2} - C B^{2}}{A^{2} - C}$
Now if we further assume that $E = B^{2} (2 - t \frac{A^{2}}{C})$ we miraculously (yes, that's sarcasm) end up with
${(A y + B)}^{2} = E - B^{2} .$
So to summarize, this "methods" works only if
$A \neq 0$

$D = \frac{2 C B}{A}$

$A^{2} \neq C$

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