This is the problem I'm having trouble with right now and its racking my brain, any help would be appreciated:

Consider a quadratic function Q(y)=Ay2+By+C. If Q(2)=0, Q(3)=0 and Q(1)=-4, find C the constant term in Q.

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- Sep 5th 2008, 11:14 AMforkball42help with finding constants in quadratice function
This is the problem I'm having trouble with right now and its racking my brain, any help would be appreciated:

Consider a quadratic function Q(y)=Ay2+By+C. If Q(2)=0, Q(3)=0 and Q(1)=-4, find C the constant term in Q. - Sep 5th 2008, 11:29 AMSimplicity
What are those boxes ()? Your question appears is appearing as Q(y)=Ay2+By+C. Is it $\displaystyle Q(y) = Ay^2 + By + C$?

For this question, substitute $\displaystyle y=2, 3, 1$ and you will get three equation which you can use simultaneous equation technique to work out constant value of $\displaystyle A, B$ and $\displaystyle C$. - Sep 5th 2008, 12:02 PMShyam
$\displaystyle Q(y)=Ay^2+By+C$

for Q(2)=0,

$\displaystyle 0=A(2)^2+B(2)+C$

4A + 2B + C =0 .......................................eqn. (1)

for Q(3)=0,

9A + 3B + C = 0 ..................................... eqn.(2)

for Q(1) = - 4,

A + B + C = - 4 .......................................eqn.(3)

subtract eqn(3) from eqn(1) and eqn (2), we get:

3A + B = 4 ............................................EQN(4)

and 8A + 2B =4

or,

4A + B =2 ..........................................EQN(5)

Solve EQN(4) AND EQN(5):

we get:

A = - 2, B = 10

Put these values in eqn(3), we got C = - 12