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HOW DO YOU SOLVE THIS QUADRATIC EXPRESSION?

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Stimshieme
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16 Oct 2008, 2:27 pm

Show that the function

f(x)=3x^2 - 6x+2

can be written as:

a(x+b)^2+c

How do I do this? Please explain and give an answer. I have a test on monday!



Ishmael
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16 Oct 2008, 2:33 pm

Wait, how does somebody just telling you the answer help if you have a test?


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Tracker
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16 Oct 2008, 2:53 pm

f(x)=3x^2 - 6x+2

f(x)=3(x^2 - 2x)+2

f(x)=3(x^2 - 2x+1-1)+2

f(x)=3(x^2 - 2x+1)+2-3

f(x)=3(x^2 - 2x+1)-1

f(x)=3(x-1)^2-1

f(x)=a(x+b)^2+c

a=3, b=-1, c=-1

does that help?



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16 Oct 2008, 3:48 pm

Tracker wrote:
f(x)=3x^2 - 6x+2

f(x)=3(x^2 - 2x)+2

f(x)=3(x^2 - 2x+1-1)+2

f(x)=3(x^2 - 2x+1)+2-3

f(x)=3(x^2 - 2x+1)-1

f(x)=3(x-1)^2-1

f(x)=a(x+b)^2+c

a=3, b=-1, c=-1

does that help?

Quadratics typically follow y= ax^2 +bx +c

f(x) = y, so set y to 0.


f(x)=3x^2 - 6x+2
f(x)=3(x^2 - 2x)+2 //subtract 2 and divide by 3
-2/3=x^2 -2x
-2/3=x(x-2)
-2/3=x and -2/3=x-2

That's what I get.


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Stimshieme
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16 Oct 2008, 5:30 pm

Tracker wrote:
f(x)=3x^2 - 6x+2

f(x)=3(x^2 - 2x)+2

f(x)=3(x^2 - 2x+1-1)+2

f(x)=3(x^2 - 2x+1)+2-3

f(x)=3(x^2 - 2x+1)-1

f(x)=3(x-1)^2-1

f(x)=a(x+b)^2+c

a=3, b=-1, c=-1

does that help?


The answer is correct but I don't understand the lines in bold, why did you add +1 and -1? And what is the name of this type of question? I can't find any resources on the internet about them. Please help I honestly feel like crying at this type of equation...



lau
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16 Oct 2008, 6:43 pm

http://en.wikipedia.org/wiki/Completing_the_square


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twoshots
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16 Oct 2008, 10:44 pm

Apera wrote:
Quadratics typically follow y= ax^2 +bx +c

f(x) = y, so set y to 0.


f(x)=3x^2 - 6x+2
f(x)=3(x^2 - 2x)+2 //subtract 2 and divide by 3
-2/3=x^2 -2x
-2/3=x(x-2)
-2/3=x and -2/3=x-2

That's what I get.

Not only was she not trying to solve it, I really don't know what you're doing here. It can readily be shown that -2/3 is not a zero of this polynomial. If you we're going to solve it, you'd probably still want to complete the square and go from there.

As for the OP, yes, this is a completing the square type problem, which is the usual way to convert ornery quadratics into a squared term plus a constant.


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