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Quadratic Function 
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#1
Nov2807, 08:40 PM

P: 135

1. The problem statement, all variables and given/known data
solve (f(x)=3 x^2 + 9 x + 1/4 ) fx=(3)(x^23 x)+1/4 fx=(3)(x^23 x+9/4)+1/4+27/4 fx=(3)(x^23 x+9/4)+28/4 fx=(3) (x3/2)^2+7 2. Relevant equations 3. The attempt at a solution the solution is already given in the book, but I don't understand why 9/4 was inserted in the third line (bolded). I know at the end of the equation, 27/4 is added to balance out the 9/4.......but why put in 9/4 there? 


#2
Nov2807, 08:52 PM

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What they've done is simply completed the square. If you have an equation of the form: [itex] x^2+2ax [/itex], you can write it as: [itex] x^2+2ax = (x+a)^2  a^2 [/itex].



#3
Nov2807, 10:25 PM

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f(x)=3 x^2 + 9 x + 1/4
they just put 27/4 and 27/4 to make completion easier f(x)=3 x^2 + 9 x 27/4+ 1/4 +27/4 =3(x^23x+9/4)+7 =3(x3x/2)^2 +7 expand (x3x/2)^2 and check it yourself 


#4
Nov2807, 10:28 PM

P: 135

Quadratic Function
yea bro...
the thing is, how do you come up with 9/4 from to make it easier? I'm on purplemath.com right now, trying to figure it out without bothering you guys... hopefully I can figure this one out. 


#5
Nov2807, 10:36 PM

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Well heres how I would have done that problem. You're given:
[tex] f(x) = 3x^2 + 9x + \frac{1}{4} [/tex] [tex] f(x) = 3( x^2 3x) + 1/4 [/tex] Now we notice that the term in brackets is of the form [itex] x^2+2ax [/itex], with [itex]a=\frac{3}{2}[/itex], and so we complete the square [itex] x^2+2ax = (x+a)^2  a^2 [/itex]: [tex] f(x) = 3\left(\left(x\frac{3}{2}\right)^2  \frac{9}{4} \right) + \frac{1}{4} [/tex] [tex] f(x) = 3\left(x  \frac{3}{2}\right)^2 + \frac{27}{4} + \frac{1}{4} [/tex] [tex] f(x) = 3\left(x  \frac{3}{2}\right)^2 + 7 [/tex] 


#6
Nov2807, 10:58 PM

P: 135

okie.........i got it.
so the point of adding 9/4 inside the brackets is so that you can make it factorable. And since you added a (), now you have to minus it to even it out. thnkx.....can't believe it took me so long. anyhow, how do you type the math out like that? 


#7
Nov2907, 02:56 AM

P: 8

You use LaTeX. This tread explains in more detail.
http://www.physicsforums.com/showthread.php?t=8997 


#8
Nov2907, 02:16 PM

P: 135

test:
[tex]x=3 x^2+9x+1/4[/tex] [tex]fx=(3)(x^2┤3 x)+1/4\\[/tex] [tex]fx=(3)(x^2┤3 x+9/4)+1/4+27/4\\[/tex] [tex]fx=(3)(x^2┤3 x+9/4)+28/4\\[/tex] [tex]fx=(3) (x┤3/2)^2+7\\[/tex] 


#9
Nov2907, 02:21 PM

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http://at.org/~cola/tex2img/index.php 


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