Hi, this is Presh Talwalkar
Quadratic equations are extremely important in mathematics
But unfortunately they can also be very confusing
In this video I'm going to present a visual and geometric proof
so that you understand the quadratic formula intuitively.
Imagine you want to solve x^2 + 2x -15 = 0
You could solve this by graphing the equation x^2 + 2x -15
The point at which the graph crosses the x axes are the solutions.
You could also solve this equation by factoring into linear terms
If you set these linear terms equal to zero, you get the same solutions as written in the graph.
When those methods don't work
The quadratic formula will always tell you how you can solve any quadratic equations with arbitrary constants a, b and c.
So this is a very useful formula
But unforunately, it's not often presented in an intuitive way.
Why exactly does it work ?
So let's try and understand this geometrically.
Let's solve this equation
First we'll add 15 to both sides so that we have x^2+2x=15
I'll now color code each term
Each term is going to represent the area of the shape.
For x^2, I'll represent it by a square, which has a side length of x.
Thea area of the square is equal to x^2.
The term 2x I'll represent by a rectangle which has a side length of 2 and another side length of x.
The area of this rectangle is equal to 2x.
Finally, the term 15 I'll represent by a rectangle which has an area of 15.
It doesn't matter what the exact side lengths are,
It only matters that the area is 15, so I'm not going to write in the side lengths.
But the task we have is: we want to have the area of the 2 shapes on the left to be equal to the area of 15, which is the shape on the right.
So how can we do that ?
The first thing we want to do is: we want to combine the two shapes.
And luckily for us, we have one common side of x.
So we can combine these two shapes on one common side.
But there's a trick we're going to use
We're going to split the rectangle of 2x into 2 small rectangles which are x by 1
And now, we're going to join these strips on each side of the square.
So now we have a shape that's almost equal to a square.
So what we need to do is we want to add on the little area so that we get a square.
So we put an orange rectangle here
An each of it's side lengths is equal to one,
And so this area is equal to one.
And in order to balance the equation, we want to add the same area to the right end side.
So what we've done geometrically, we'll represent algebraically by adding a term of one to each the left and the right end sides of the equation.
And this is what's known as "completing the square"
You probably heard the term in algebra class, but this is the geometric reason.
We are literally adding a term to this area so that we can complete a square.
And now this helps us solve the problem.
Because what we want to do was simplify that 15+1=16
We have a square with the side lengths of x+1
and we need that square to be equal to the square on the right hand side.
So we want (x+1)^2 to have an area of 16
Now one way we could to this, to have this area of 16,
is we can imagine to have a square that has a side length of 4
So in order for our square on the left to be equal to the square on the right,
we want x+1 = 4
We can easily solve this, x=3
But there's another way we can do this.
It's kind of an usual way, but imagine we had side length of (-4)
Since (-4)x(-4) = 16
So we can also set (x+1) = -4
So another solution is x= -5
So this is the visual way to understand quadratic equations.
And if you understood this, I'm going to show you this proof, just in a more general way,
which is how we get the quadratic formula.
So we have a general quadratic equation
And first I'm going to subtract "c" from both sides
so we have ax^2+bx = -c
we'll now divide by the constant term of a, so we have x^2+(b/a)x = -c/a
This is very similar to the equation we just solved.
It is: x^2 + (something)*x = constant
So let's use the same procedure
We'll represent x^2 as being the area of the square with the side length of x.
We'll represent (b/a)x by a rectangle which as one side of (b/a) and another side of x.
And we'll represent (-c/a) by a rectangle which as an area of (-c/a)
We don't need to write the lengths of the sides, we just know that the area is (-c/a).
So we want the area of the two shapes on the left to equal the area of the shape on the right.
We'll use the same trick as in the last problem.
We'll divide our middle green rectangle in half, so that we have two side lengths of b/(2a)
We'll now combine that with the square which has a side length of x, by matching the sides of x.
And now we almost have a square
What we're missing is an orange square
and each of it's side lengths is b/(2a)
So b/(2a)*b/(2a) = (b^2)/(4a^2) gives us the area of the orange square
In order to balance the areas, we have to add the same area to the right end side.
So now, let's do this algebraically.
We need to add b^2/(4a^2) to both sides of the equation.
We'll simplify the right hand side of the equation
as we combine the two like terms, I'm going to skip a couple of steps,
But the right end side becomes (b^2-4ac)/(4a^2)
So that becomes the area of the rectangle on the right hand side.
The left hand side we have a square with a side length of x+b/(2a)
And we want that to be equal to our square, which has an area of (b^2-4ac)/(4a^2).
So there's two way we could do this:
We could set this equal to positive the square root
or negative the square root of the area of the right hand side
So we have the side length on the left , x+b/(2a), will either equal to the positive root or the negative root
And now if we simplify this square root
And then move term b/(2a) to the right hand side
we end up with the quadratic formula that x=((-b)+/- sqrt(b^2-4ac))/(2a)
And that's where the quadratic formula comes from intuitively
as that's this is the length
that will allow the square on the left to be equal to the square on the right.
So i hope this gives a deeper understanding of the quadratic formula
and its geometric interpretation.
Thanks for watching this video, please subscribe to my channel
I make videos on math and game theory
you can catch me on my blog
and facebook, google + and patreon
yon can catch me on social media @preshtalwalkar
And you can get my books, there are links in the video description.