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spaminator101
08-22-2005, 07:52 PM
i first posted this in science math and philosophy
they told me to post it here

"so im studying quadratic functions and need help solving quadratic equations by completing the square
im just not getting it"

BruceZ
08-22-2005, 09:01 PM
[ QUOTE ]
i first posted this in science math and philosophy
they told me to post it here

"so im studying quadratic functions and need help solving quadratic equations by completing the square
im just not getting it"

[/ QUOTE ]

Say you have the quadratic 4x^2 + 12x - 7 = 0.

First divide through by 4 to get x^2 by itself:

x^2 + 3x - 7/4 = 0

x^2 + 3x = 7/4

Now the trick is to add something to the left hand side so that we can write it as (x + c)^2. That would be x^2 + 2cx + c^2. We already have x^2 + 2cx where c = 3/2, so we add (3/2)^2 = 9/4 to both sides since x^2 + 3x + 9/4 = (x + 3/2)^2. In general, the magic number that we add is always going to be the coefficient of x divided by 2 and then squared. Remember to make sure the leading coefficient of x^2 is 1 first.

x^2 + 3x <font color="blue">+ 9/4</font> = 7/4 <font color="blue"> + 9/4</font>

(x + 3/2)^2 = 4

x + 3/2 = 2 or x + 3/2 = -2

x = 1/2 or x = -3.5

The whole trick is knowing what to add to each side to make this work. In general, for x^2 + bx + c = 0, the magic term is always (b/2)^2. So you take half the coefficient of x and square it. Don't forget to add it to both sides.

That's how we derive the quadratic formula:

ax^2 + bx + c = 0

x^2 + (b/a)x + c/a = 0

x^2 + (b/a)x = -c/a

x^2 + (b/a)x <font color="blue">+ (b/2a)^2</font> = -c/a <font color="blue">+ (b/2a)^2</font>

(x + b/2a)^2 = -c/a + (b/2a)^2

x + b/2a = +/- sqrt[-c/a + (b/2a)^2]

x = -b/2a +/- sqrt[(-4ac + b^2)/4a^2]

x = [-b +/- sqrt(b^2 - 4ac)] / 2a

spaminator101
08-22-2005, 10:08 PM
and so basicly you just use the quadratic formula to complete the square

if so this isnt so hard after all

spaminator101
08-22-2005, 10:13 PM
should i move this part of the discussion to the s.m.p. forum

BruceZ
08-22-2005, 10:53 PM
[ QUOTE ]
and so basicly you just use the quadratic formula to complete the square

if so this isnt so hard after all

[/ QUOTE ]

No, the quadratic formula is another method you can use to solve any quadratic equation, besides completing the square. It's just that we derived the quadratic formula by completing the square for a general quadratic. Then now that we have that formula, we can just use that instead of completing the square manually every time, but your instructor may ask you to solve by completing the square, so then you would have to do it manually as I did in the first example, instead of using the formula.

We can just continue the discussion here.

spaminator101
08-23-2005, 03:47 PM
ok ive got it now
thanks brucez

DiceyPlay
08-24-2005, 11:31 AM
Use completing the square to derive the quadratic formula.

end with x = (-B (+-) SQRT(B^2-4AC))/2A

The algebraic manipulation is a little tricky. This goes a long way in solidifying your algebraic understanding and skills in my opinion.

-DP

BruceZ
08-24-2005, 04:02 PM
[ QUOTE ]
Use completing the square to derive the quadratic formula.