at a basic GCSE level, i am stuggling with this concept
(3x-b)^2 - 9 = ax^2-12x+c
find a,b and c for all values of x
i was told taht
9x^2 -6xb+b^2 - 9=ax^2-12x + c
therefore, a must equal 9 b must equal minus 2
this emans that the equation falls nicely into place
then when x=0
b^2 - 9=c
substituting that in
however, i argued that there was more than one solution and actually, becuase there is 4 variables, and only 1 equation, it is impossible to figure all the equation out with numbers
e.g c could equal anything, and this would balance out any differences, meaning that a,b or c could be anything
why am i wrong?
a can only be 9 because the x² must match on both sides since you have an equal sign, thus b can only be 2 due to the same reason with x, which reduces the number of variables to 2. On the left side of the equation you still have b²-9, which should be c according to the right side. With the already known b=2, c becomes -5. So the only variable you got left is x.
a, b and c can only equal 9, 2 and -5 respectively, anything else wouldn't solve the equation. This is if a, b and c is not a function of x, of course, and just numbers.
Bang postponed. Not big enough. Reboot.
thanks, you have cleared things up alot, so it works, as long as each one is an actual number and not a function of x, that was the thing that kept catching me out, i simply couldn't see how it worked, but the assumption has to be made that it is not a function of x