maths help plz?

Question

Solve for a & b ìf 2x²-12x+9=2(x-a)²+b

Another one:
The expression 3x²-x-9 is rearranged into the form:
a(x+2)²+b(x+2)+c. What are the values of a,b and c?

I would appreciate explanations and working.

Answer 1

open the brackets will make it look simpler

2x²-12x+9=2(x-a)²+b
2x²-12x+9=2x²-4ax+2a²+b

by comparing coefficients of x,
-12=-4a
a=3

by comparing coefficients of x^0,
9=2a²+b
9=18+b
b=-9

next qn:
3x²-x-9=a(x+2)²+b(x+2)+c
3x²-x-9=ax²+(4a+b)x+2b+c+4a

by comparing coefficients of x², a=3

by comparing coefficients of x,
-1=4a+b
-1=12+b
b=-13

by comparing coefficients of x^0,
-9=4a+2b+c
-9=12+(-26)+c
c=5

there you go!(:

Answer 2

2x² - 12x + 9 = 2(x-a)² + b
2x² - 12x + 9 = 2x² - 4ax + 2a² + b

By comparing coefficients of x,
-12=-4a
a=3

By comparing coefficients of x^0 i.e constant,
9 = 2a² + b
9 = 18 + b
b = -9

NEXT PROBLEM

3x² - x - 9 = a(x+2)² + b(x+2) + c
3x² - x - 9 = ax² + (4a+b)x + 2b + c + 4a

Compare coefficients of x²,
a=3

Compare coefficients of x,
-1 = 4a + b
-1 = 12 + b
b = -13

Compare coefficients of x^0 i.e Constant term,
-9 = 4a + 2b + c
-9 = 12 + (-26) + c
c = 5

Answer 3

i solve your problem by min of the function:
if x=a......y=b and notice( b) is the min of the function
in the other hand you have the x(min):
x=-(-12)/2x2=3=a
put(3)in (x)and you will see:
2x9-12x3+9=18-36+9=-9
for your other question:do not forget to comparison=ax*2=3x*2 so:a=3
x=-2 is the x(min)so put it in(3x*2-x-9).you see y=5=c(do not forget the max and min of the cubic function)
now for finding b: use x=0.....12+2b+5=-9...2b=-26...b=-13
sorry my English was not very good