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andieje

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completing the square of quadratic equation

Hi

If you have a quadratic iequation in the form ax^2 + bx + c = 0, I know how to complete the square to get the well-known formula which I won'r type out.

I have just read this and I don't understand how to complete the square in this different context:

for f(x) = ax^2 + bx + c we can complete the square on the right hand side and simplify to get....
 
I don't know how to complete teh square in this case. This is the only way i know how to do it:

ax^2 + bx + c = 0  --> take c to other side of equation
ax^2 + bx = -c ---> remove the coefficient of x^2

I won't go though all the steps but you get the picture. How do i complete the square in this case:

f(x) = ax^2 + bx + c

Perhaps if you show me the first few steps it will be obvious and I can complete the rest of it.

Thanks















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MYCU
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I think you should complete the square to factor the expression rather than solve it.
The steps are essentially the same.


The Quadratic Formula give     x  =  -b  +/- sqrt(b² - 4ac)/2a


In this case you should wind up with something like

     f(x)  =  ax^2 + bx + c  =  (x + b  + sqrt(b² - 4ac)/2a) * (x + b  - sqrt(b² - 4ac)/2a)
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andieje

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The answer is supposed to be

f(x) = (4ac - b^2)/4a + a(x + b/2a)^2

I'm not sure how it got that.
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MYCU
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I do not understand how the contexts are different? You just complete the square as usual on the RHS.


(4ac - b²/4a + a(x + b/2a)² is merely the completion of the square of   ax²  + bx + c   ie that


 ax²  + bx + c  = (4ac - b²/4a + a(x + b/2a)²


so that

if               f(x)=  ax²  + bx + c

then           f(x)= (4ac - b²/4a + a(x + b/2a)²

SOLUTION
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ASKER

Can i do this then as the first step:

f(x)=  ax²  + bx
f(x) - c = ax^2 + bx

I'm not that comfortable with function notation. Not used it for years. I guess its valid to write f(x) - c?
How do you get the sqaured notation glynweb? f(x)=  ax²  + bx
I ahve to type ^2
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ASKER

sorry, of course its the same. It looked completely different on first glance but if you work it through its the same. I also thought it had to equal 0 for you to do this.

thanks