How do you do quadratic equations? What is the quadratic formula and what is the best way to remember it?

1 Answer
Jun 22, 2018

See explanation

Explanation:

color(blue)("Formula type 1")

When I was in my upper years of school I made a point of writing down the quadratic formula each time I needed to use it. The repetition of this act really fixed it into my memory. There was a lot of repeats and it is the repetition of ul("writing it down") that did the trick. I am a lot older and I can still remember it.

Given the standard form of y=ax^2+bx+c then if you set y=0 we have for that condition: x=(-b+-sqrt(b^2-4ac))/(2a)

As long as the part b^2-4ac is positive and greater than 0 the graph crosses the x-axis ( 2 values for x).

If the part b^2-4ac = 0 then the x-axis forms a tangent to the vertex.

If b^2-4ac is negative then the graph does not cross the x-axis and you are entering the realm of what is called complex numbers.

If the part ax^2 is positive then the graph is of general shape uu

If the part ax^2 is negative then the graph is of general shape nn
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color(blue)("Formula type 2")

There is another format which has the same values as the previous one (on the whole) and again you need to repeatedly write this one down. It is the Vertex equation.

Given y=ax^2+bx+c" then "y=a(x+b/(2a))^2+k_1+c

We need k_1 as changing into this format introduces a value that is not in the type y=ax^2+bx+c. So we remove this by setting

k_1+a(b/2a)^2 =0 We then add this to c giving k_2:

y=a(x+b/(2a))^2+k_2" where "k_2=k_1+c

x_("vertex")->(-1)xx(b/(2a)) = -b/(2a)
y_("vertex")=k_2

Determine the x-intercepts you set y=0=a(x+b/(2a))^2+k_2 and solve for x.