Converting a perfect square to quadratic format? (6z+2z)^2 = (ax+2z+8)(-8+bx+cz)

[danielw]

Hi all

Newbie here, so please be kind.

\(\displaystyle \mbox{(a) Find }\, a,\, b,\, c\, >\, 0\, \mbox{ so that:}\)

. . .\(\displaystyle (6x\, +\, 2z)^2\, -\, 64\, =\, (ax\, +\, 2z\, +\, 8)\, (-8\, +\, bx\, +\, cz)\)

I have this problem, above. It seems to be asking me to convert from a completed square format to the quadratic format, but has given some co-efficients. Is that the correct wording?

Steps taken so far:

= 36x^2+24xz+4z^2-64 [I expanded out the equation]

But after this I don't know what technique to follow.

I hope you can help!

Thanks

Daniel

 

[Deleted member 4993]

Hi all

Newbie here, so please be kind.

\(\displaystyle \mbox{(a) Find }\, a,\, b,\, c\, >\, 0\, \mbox{ so that:}\)

. . .\(\displaystyle (6x\, +\, 2z)^2\, -\, 64\, =\, (ax\, +\, 2z\, +\, 8)\, (-8\, +\, bx\, +\, cz)\)

I have this problem, above. It seems to be asking me to convert from a completed square format to the quadratic format, but has given some co-efficients. Is that the correct wording?

Steps taken so far:

= 36x^2+24xz+4z^2-64 [I expanded out the equation] ... not needed

But after this I don't know what technique to follow.

I hope you can help!

Thanks

Daniel

use:

m² - n² = (m + n)(m - n) noting 64 = 8²