Why Was 9/4 Added in the Quadratic Equation Completion?

[viet_jon]

[SOLVED] Quadratic Function

Homework Statement

solve (f(x)=-3 x^2 + 9 x + 1/4 )
fx=(-3)(x^2-3 x)+1/4

fx=(-3)(x^2-3 x+9/4)+1/4+27/4

fx=(-3)(x^2-3 x+9/4)+28/4
fx=(-3) (x-3/2)^2+7

Homework Equations

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?

 

[Kurdt]

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

 

[rock.freak667]

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^2-3x+9/4)+7
=-3(x-3x/2)^2 +7

expand (x-3x/2)^2 and check it yourself

 

[viet_jon]

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.

 

[Kurdt]

Well here's how I would have done that problem. You're given:

$$f(x) = -3x^2 + 9x + \frac{1}{4}$$
$$f(x) = -3( x^2 -3x) + 1/4$$

Now we notice that the term in brackets is of the form $x^2+2ax$, with $a=\frac{-3}{2}$, and so we complete the square $x^2+2ax = (x+a)^2 - a^2$:

$$f(x) = -3\left(\left(x-\frac{3}{2}\right)^2 - \frac{9}{4} \right) + \frac{1}{4}$$
$$f(x) = -3\left(x - \frac{3}{2}\right)^2 + \frac{27}{4} + \frac{1}{4}$$
$$f(x) = -3\left(x - \frac{3}{2}\right)^2 + 7$$

 

[viet_jon]

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?

 

[jaime2000]

You use LaTeX. This tread explains in more detail.

https://www.physicsforums.com/showthread.php?t=8997

 

[viet_jon]

test:

$$x=-3 x^2+9x+1/4$$

$$fx=(-3)(x^2┤-3 x)+1/4\\$$

$$fx=(-3)(x^2┤-3 x+9/4)+1/4+27/4\\$$

$$fx=(-3)(x^2┤-3 x+9/4)+28/4\\$$

$$fx=(-3) (x┤-3/2)^2+7\\$$

 

[Kurdt]

test:

<tex>x^2</tex>

<tex>(x)=-3 x^2+9x+1/4</tex>
fx=(-┤3)(x^2┤-3 x)+1/4
fx=(-┤3)(x^2┤-3 x+9/4)+1/4+27/4
fx=(-┤3)(x^2┤-3 x+9/4)+28/4
fx=(-┤3) (x┤-3/2)^2+7

If you want to test then there are plenty of preview websites. Here is one of them.

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