perfect square

suvadip

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Joined
Feb 21, 2012
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7
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
 
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!
 
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
 
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