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**thinkdesigns****Member**- Registered: 2005-07-21
- Posts: 23

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

-5=c

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?

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**numen****Member**- Registered: 2006-05-03
- Posts: 115

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.

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**thinkdesigns****Member**- Registered: 2005-07-21
- Posts: 23

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

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**numen****Member**- Registered: 2006-05-03
- Posts: 115

If a, b or c is a function of x, it is usually written a(x), b(x) or c(x).

Bang postponed. Not big enough. Reboot.

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