Algebra 1 Tutorial
Where did we get the Quadratic Formula?
This tutorial derives the Quadratic Formula to show students where it came from.
Solve for the roots of the quadratic equation:
y = ax^2 + bx + c
Let’s say we have a generic quadratic equation using variables a, b, and c:
y = ax^2 + bx + c
Now, given this equation, how would you solve for the roots (I.e. the values of x where y = 0)?
First, set y to 0.
0 = ax^2 + bx + c
Now, we’re trying to isolate x to one side so that we get x = (some values).
First, move c to the other side by subtracting it from both sides.
-c = ax^2 + bx
Now we’ll divide both sides by a, to get the a out of that x^2 part of the equation.
-c/a = x^2 + (b/a)x
Now we have to figure out how to get ‘x’ by itself.
It almost looks factorable, but there is a missing variable. We’ll call this variable ‘d’.
-c/a + d^2 = x^2 + (b/a)x + d^2
Now we have to find out what d is for the right side of the equation to be able to factor.
(x+d)^2 = x^2 + 2dx + d^2
So… looking at the middle term of the right side of the equations
b/a = 2d
b/2a = d
Plugging in ‘d’, we see that:
(x+(b/2a))^2 = x^2 + (b/a)x + (b/2a)^
= x^2 + (b/a)x + (b^2)/(4a^2)
So now we have:
-c/a + (b^2)/(4a^2) = x^2 + (b/a)x + (b^2)/(4a^2)
Since we know that the right side factors nicely, we can go ahead and do that.
-c/a + (b^2)/(4a^2) = (x+(b/2a))*(x+(b/2a))
-c/a + (b^2)/(4a^2) = (x+(b/2a))^2
Now we can take the square root of both sides to get closer to getting x by itself.
±√(-c/a + (b^2)/(4a^2)) = x+b/2a
Note that when you take the square root of the left side, a + as well as – term is produced (i.e. -1^2 = 1 and 1^2 = 1 as well).
Now, moving the b/2a term to the other side, we get:
-b/2a ± √(-c/a + b^2/(4a^2)) = x
This may look funny, but after some simple manipulation we have come to the quadratic equation!
First get a common denominator inside the square root.
-b/2a ± √(b^2/(4a^2) – 4ac/(4a^2))
Now since the 4a^2 is inside a square root, and is also a perfect square, we can factor it out of the square root function by taking the square and putting it under the square root.
-b/2a ± √(b^2 – 4ac)/±2a
Note: the ± by the 2a was created from square rooting 4a^2, but since there is already a ± it kind of gets ignored (i.e. the top of the function could be – or + and so could the bottom, but if both are negative the whole thing is positive, and any combination of the two can only lead to either positive or negative values).
[-b ± √(b^2 – 4ac)]/2a
The quadratic formula was born! This means that given any values of a, b, and c, one can easily solve any combination of quadratic equations!
About The Author
|I am skilled in the areas of mathematics as well as computer technologies. I have been excelling at mathematics since I was in 5th grade. I started UIL Number Sense in Texas when I was in 5th grade, and from there I have expanded my knowledge of mathematics greatly. I started college when I was just...|