perfect square

[suvadip]

For which value of y(any real number),

(16+y)x²-4(1+y)x+(y²-128)/4 is a perfect square?

Please help me

 

[soroban]

Hello, suvadip!

$\displaystyle \text{For real value of }y\text{ is }\,(y+16)x^2 - 4(y+1)x - \dfrac{y^2-128}{4}\,\text{ a perfect square?}$

A quadratic is a square if its discriminant, $\displaystyle b^2\!-\!4ac$, is zero.

$\displaystyle D \;=\;[4(y+1)]^2 - 4(y+16)\left(\frac{y^2-128}{4}\right)$

. . .$\displaystyle =\;16(y^2+2y+1) - (y+16)(y^2-128)$

. . .$\displaystyle =\;16y^2 + 32y + 16 - (y^3 + 16y^2 - 128y - 2048)$

. . .$\displaystyle =\;16y^2 + 32y + 16 - y^3 - 16y^2 + 128y + 2048$

. . .$\displaystyle =\;-y^3 + 160y + 2064$


$\displaystyle \text{If }y^3 - 160y - 2064 \:=\:0,\, \text{ the only rational roots are:}$ \(\displaystyle 1,2,4,6,8,16,18,24,\,\hdots\)

$\displaystyle \text{If }y = 16,\,\text{ we get: }\,-528$
$\displaystyle \text{If }y = 18,\,\text{ we get: }\,+888$

There is an irrational root between 16 and 18.

. . Good luck!

 

[suvadip]

Perhaps I was unable to make you understand. I need only the real value of y for which the given expression will be a perfect square. x should be present in the expression. I want to write the expression as (ax-b)² . Actually I want to solve x⁴-8x³+4x+32=0 by Ferrari method. Thats why I need such y.

Huh? Is that the FULL original problem?

For now, the expression is a perfect square in infinite cases; like:
x=0, y=12 : 4
x=0, y=18 :49
x=1, y=36 : 196
x=2, y=6 : 9