MHB Find unknowns in equation going from perfect square? to quadratic format?

[danielw]

Hi all

I'm trying to work out how to answer this type of problem.

$$(6x+2z)^2-64=(ax+2z+8)(-8+bx+cz)$$ where a, b and c > 0

I have attempted the problem by expanding the brackets:

$$=36x^2+24xz+4z^2-64$$

This is the same as $$(6x+2z)^2-(8)^2$$

Then subtracting from either 'side' of the quadratic and multiplying the result:

$$=((6x+2z)-8)*((6x+2z)+8)$$

So visually comparing these results I got
$$a=6\\
b=2\\
c=2$$

But this is wrong.

I think my method may be wrong.

I'm looking for guidance on how to solve this kind of problem.

I hope one of you can help.

Many thanks

Daniel

 

[Greg]

Hi danielw and welcome to MHB! :D

The method you're using is usually called "equating coefficients" and it's a good approach. Review your work and try $a=b=6,\,c=2$.

 

[HallsofIvy]

Note that, in order to be able to solve for a, b, and c, the equation must be true for all x and y.

 

[soroban]

\text {Solve for }a,b,c: \;(6x+2z)^2-64\:=\: (ax+2z+8)(bx+cz-8)

\text{Expand: }\;36x^2 +24xz + 4z^2 - 64 \;=\;abx^2 + acxz - 8ax + 2bxz +2cz^2 - 16z + 8bx + 8cz - 64

\text{Equate coefficients: }\;\;36 \:=\:ab, \quad 24 \:=\:ac + 2b, \quad 4 \:=\:2c, \quad 8b-8a - 0, \quad 8c - 16 \:=\:0

\text{And we get: }\;\;\boxed{ a = 6,\;b = 6,\;c = 2}