quadratic formula??

Question

can someone show me how they come up with the quadratic formula? i dont mean the history, just show me how they come up with that formula.... thanks :)

cool thanks! :)

Answer 1

http://en.wikipedia.org/wiki/Quadratic_formula#Derivation

Answer 2

Good, good question!
The general EXPRESSION for a quadratic is
ax^2 + bx + c. When graphed, the graph cuts the x-axis at two points. These "crossing points" are called the roots of the quadratic, and at both crossing points, the y term of the ordered pair is zero, as in 0

The general EQUATION of any quadratic is
y=ax2 + bx + c . Tell me the x term and I can calculate the y term. When they ask you to solve any quadratic equation, they are really saying "find the roots of". And THAT means find the crossing points. And at the crossing points, y=0. The problem thus changes to this: "For the Quadratic equation you have been given, find what values of x make y equal to 0".

Lots of quadratics can be solved by factoring. For example if you are asked to solve x^2 - 3x + 2 = 0, you simply factor, to get (x-2)(x-1)=0. Now, for (x-2)(x-1) to multiply out to 0, one or the other or both factors has to be zero. If it's x-2 that = 0, then x=2 . If it's x-1 that=0, then x=1. You have found the roots, or crossing points, where y=0.

When factoring is a nightmare, the quadratic formula rides to the rescue. It always works.
It's development is based on completing the square, and before I launch into my windy
story, I want to review competing the square with you.
x^2 +6x +5 =0, so (x+5)(x+1)=0, x=-5 or-1

Let's factor this by completing the square.
x^2 + 6x + 5 =0
x^2 + 6x = -5
x^2 + 6x +9 = -5 +9
(x + 3)^2 = 4
(x+3) = +or- 2
x = +2-3 or -2-3, from which x=-1 or -5
The key steps in factoring by completing the square are:
1) Arrange terms in descending powers of x
2)MAKE DAMN SURE THE COEFFICIENT OF THE X^2 TERM IS +1
3)Transfer the constant term(that's the term with no x attached to it) to the right-hand side of the equation.
4) Make the left hand side a perfect square. This is done by taking HALFof the coefficient of the x term and squaring it. Be sure to add the same amount to the right hand side of the equation, so as to maintain the value of the equation.
5) Take the square root of both sides
6) Isolate x.

OK, here we go!

ax^2 + bx +c =y
ax^2 +bx + c = 0 (we're finding roots)
ax^2 + bx = -c
x^2 + (b/a)x = -c/a
x^2 + (b/a)x +b^2/4a^2 = -c/a + b^2/4a^2
x^2 + (b/a)x + b^2/4a^2= -4ac + b^2
nothing here. to keep my ---------------
computer from messing 4a^2
things up

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



x + b/2a = + or - sq rt (b^2 - 4ac)/2a



TA DAH!!!!

x = -b + or -sq rt (b^2 - 4ac)
----- ------------------------------
2a 2a

x= -b +or- sq rt (b^2 - 4ac)
----------------------------------
2a

The term under the square root sign is called the DISCRIMINANT. IF it's positive, you will have 2 real and different roots. If it's zero you have two roots, both the same. If it's negative, you cannot take the square root of a negative number and get a number that belongs to the Real Number System. We say no real roots.

I truly hope this is of some help to you. I have always admired students who have the courage to ask WHY? Good luck to you

Answer 3

let ax^2 + bx + c = 0
dividing through by a,
x^2 + (b/a)x + c/a) = 0
completing the square,
x^2 + (b/a)x + (b/2a)^2 - (b/2a)^2 + c/a = 0
(x^2 + (b/2a))^2 - b^2/4a^2 + 4ac/4a^2 = 0
(x^2 + (b/2a))^2 - (1/4a^2)(b^2 - 4ac) = 0
factoring the difference of squares,
(x + b/2a - √[(1/4a^2)(b^2/4a^2 - 4ac/4a^2)]) (x + b/2a + √[(1/4a^2)(b^2/4a^2 - 4ac/4a^2)]) = 0
x = - b/2a - (1/2a)√(b^2 - 4ac), - b/2a + (1/2a)√(b^2 - 4ac)
x = - b(1/2a) ± (1/2a)√(b^2 - 4ac)
x = (1/2a)[- b ± √(b^2 - 4ac)]
x = [- b ± √(b^2 - 4ac)] / 2a

Answer 4

x=-b+ or - square root of b^2 - 4ac
all divided by 2a

i just memorized it i guess.... what do you mean by how they came up with it?