What is the quadratic equation and what does it do?
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When you chuck something it follows a path called a parabola. Say you throw a ball, now it’s going 12 meters per second up and 15 meters per second sideways and you throw it from 2 meters off the ground and you want to calculate where it will land. Say gravity is -10 m/s^2, that is: every second in the air the ball accelerates down with 10 more meters per second of downward speed. That’s all gravity is.
Now I’m going to start running through some algebra, you don’t need to understand every step but I want you to understand where the letters come from to understand why the quadratic formula has so many letters in it.
First we need to write the ball’s position as algebra.
Horizontally the ball just keeps going sideways at 15 m/s. Gravity can’t change horizontal speed.
So x is the ball’s position horizontally and t is seconds. x=15t is how we say its horizontal position is just 15 * the number of seconds since you threw it
Now to find the balls height over time. So y is the ball’s height and t is seconds: y = -10*t*t + 12*t + 2
Which is how you write that the ball is accelerating down at 10m/s^2 and starts out at 12 m/s of speed going up and starts 2 meters off the ground.
But we want to know *where* the ball lands not *when* so we need to get rid of seconds and just work in meters. Algebra helps us do that. Algebra tells us that if x=15t then t=x/15
And we use this to replace t with x/15. That means we’re replacing time with a formula for horizontal distance
y = -10*t*t + 12*t + 2 becomes y = -10*x/15*x/15 + 12*x/15 + 2
If you know some more algebra that simplifies to y = -(2/45)*x^2 + (4/5)*x + 2
Now we have an equation that shows the path of the ball. We can put in x (horizontal distance) and get y (vertical height) and you can graph this and see a real path of the ball.
Now we want to find where the ball lands. You could guess and check. Try out values of x and see how close you can get y to be 0 but we have an equation: the quadratic formula that solves it for us.
You need an equation in the form y=a*x^2 + b*x + c and it will tell you where it crosses y=0. For us that means where it hits the ground.
Our equation is in that form so a=-2/45, b=4/5, and c=2
We plug it all into that big ugly equation and we get two answers?? One says the ball lands at 20.225 meters after we threw it and the other answer says it lands -2.225 meters behind us.
There is a plus or minus in the quadratic equation because parabolas cross y=0 twice. In this case the full parabola crossed y=0 before we even threw it.
Parabolas can also just not cross y=0 at all, like if you threw a ball from under zero and the throw just didn’t make it up to the roof where zero is. In this case the quadratic formula will give you two square roots of negative numbers or two “imaginary” numbers. For now you can throw these out and just say it never crosses y = 0
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