## Algebra 1 Tutorial

#### Intro

This tutorial derives the Quadratic Formula to show students where it came from.

#### Sample Problem

Solve for the roots of the quadratic equation:

y = ax^2 + bx + c

[b±√b^2-4ac)]/a

[-b±√(b^2-4ac)]/2a

[-b+√(b^2-4ac)]/2a

[b±√(b^2-ac)]/2a

[-b-√(b^2-4ac)]/2a

#### Solution

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))

or…

-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).

Therefore:

[-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! 