completing the square?

Question

a) express 3x^2 - 12x + 5 in the form A (x - B)^2 - C , where A, B and C are constants whose values should be stated

b) find the minimum value in
I) 3x^2 - 12x + 5
II) (3x^2 - 12x + 5) ^2

c)for each expression in b state a value of x which gives the minimum value

Answer 1

To complete the square 3x^2 - 12x +5

1) find the value of A - just the number in front of the x^2 term and factor out of the x^2 and x value
3x^2 - 12x +5
=3(x^2 - 4x) +5

2) find the value of B - equal to half the number now in front of the x term (after A factored out) - need to remember to take away an extra A*(B^2) to find C at the end.
=3(x^2 - 4x) +5
=3(x^2 - 4x) +(12 - 12) +5
=3(x^2 -4x +4) -12 +5
=3(x-2)^2 -7

So A=3, B=2 and C=-7

This is a y=x^2 graph which has been shifted 7 units down and 2 units to the right.

=> minimum value of 3x^2 - 12x + 5 becomes -C = -7
which occurs at x=B=2


If a graph is squared then all the negative parts will reflect back up over the x axis (since can't have square being negative) - and the steepness of the graph will also increase.

So the minimum value of [f(x)]^2 is either the [minimum of f(x)]^2 if f(x) is never negative
Or = 0 if f(x) is negative at some points.

We know f(x) was negative - since minimum=-7
So minimum of [f(x)]^2 will be = 0
This will happen at the same places that f(x)=0

So we need to solve 3x^2 - 12x + 5 = 0
=> x=[12 +/- √(144 - 60)]/(2x3)

=> x=[12 +/- √84]/6 = 2+/- (√21)/3
x=0.4725
x=3.5275

So minimum of (3x^2 - 12x + 5) ^2 is zero and occurs at x=0.4725 and x=3.5725

Answer 2

(3x^2-12x) +5=3 (x^2-4x)+5 = 3(x^2-4x +(2^2) -(2^2))+5
=3(x^2-4x+4)-3(4)+5= 3(x-2)^2 -7 hence A=3,B=3, C=7
i) minimum when x-2=0 i:e when x=2 and minimum value is -7
ii) (3x^2-12x+5)^2 = [3(x-2)^2-7]^2
the second part is hard to find minimum value and it has no connection with first part. so by taking derivative and equating to zero gives x=2, and x= 2+(7/3)^(1/2)

Answer 3

3x^2 - 12x + 5
=3(x^2-4x +5/3)
=3(x^2 -4x+4)+5-12
3(x-2)^2 -7
So A = 3, B= 2, C = 7

I) Min occurs when x=2 which is -7
II) Min is 0 at x= [6 +/- sqrt(21)]/3