SOLUTION: If 3x^2-12x+14=3(x+p)^2+q, what are the values of p and q?

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Question 266245: If 3x^2-12x+14=3(x+p)^2+q, what are the values of p and q?
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
3x%5E2-12x%2B14=3%28x%2Bp%29%5E2%2Bq
Since it must be identically true regardless of the value of x,
the easy way is to substitute any two values of x and solve the
resulting system:
Substitute x=0
3%280%29%5E2-12%280%29%2B14=3%28%280%29%2Bp%29%5E2%2Bq
14=3p%5E2%2Bq
Substitute x=1
3%281%29%5E2-12%281%29%2B14=3%28%281%29%2Bp%29%5E2%2Bq
3-12%2B14=3%281%2Bp%29%5E2%2Bq
5=3%281%2Bp%29%5E2%2Bq
So we have this system of equations:
system%2814=3p%5E2%2Bq%2C5=3%281%2Bp%29%5E2%2Bq%29
Solve both for q:
system%28q=14-3p%5E2%2Cq=5-3%281%2Bp%29%5E2%29
Set their right sides equal and solve
14-3p%5E2=5-3%281%2Bp%29%5E2
Simplify and solve this and you'll get
one solution p=-2, then substitute
that into q=14-3p%5E2 and get
q=14-3%28-2%29%5E2=14-3%284%29=14-12=2
Edwin