Re: i need some help
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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"
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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
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