So whatever happened to FOIL (First - Outside - Inside - Last), or am I just hopelessly outdated?PEMDAS for the win!
So whatever happened to FOIL (First - Outside - Inside - Last), or am I just hopelessly outdated?PEMDAS for the win!
That has to do with solving equations like (x+2)(x-3)=7. It gives you the long format of x^2-3x+2x-6=7 or x^2-x-6=7. Then you work it down to x^2-x=13, then x(x-1)=13. From there I forgot how to solve it further, been way to long lol.So whatever happened to FOIL (First - Outside - Inside - Last), or am I just hopelessly outdated?
I don't know if any of you younger fellas remember when we had math drills in grade as well as high school. I was always stumped because no one at home helped me and certainly no teacher helped me until I entered the eleventh grade and then a teacher helped me to understand math and I actually enjoyed it, I went from a D to a B. Now reading...I could read a book in one day and since I read so much, I could spell and had no problem in literature and writing a report, I could do it in class while the rest of them took the work home and worked evening on their paper. In college if a report was required during a class, I would do it first because it was so easy. But I often think of that teacher who took a little time with me and explained math. I wish I had thanked him and if I knew where his daughter lived, I'd look her up and tell her of my gratitude.
If you are trying to solve for x, put in the standard quadratic form: x^2 -x -13 = 0. (ax^2 + bx +c = 0)That has to do with solving equations like (x+2)(x-3)=7. It gives you the long format of x^2-3x+2x-6=7 or x^2-x-6=7. Then you work it down to x^2-x=13, then x(x-1)=13. From there I forgot how to solve it further, been way to long lol.
See why I hate math?If you are trying to solve for x, put in the standard quadratic form: x^2 -x -13 = 0. (ax^2 + bx +c = 0)
Then use the quadratic formula ( https://en.wikipedia.org/wiki/Quadratic_formula )
with x representing an unknown, a, b and c representing constants with a ≠ 0, the quadratic formula is:
{\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\ \ }
where the plus–minus symbol "±" indicates that the quadratic equation has two solutions.[2] Written separately, they become:
{\displaystyle x_{1}={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\quad {\text{and}}\quad x_{2}={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}}
Each of these two solutions is also called a root (or zero) of the quadratic equation. Geometrically, these roots represent the x-values at which any parabola,See why I hate math? explicitly given as y = ax2 + bx + c, crosses the x-axis.
As well as being a formula that yields the zeros of any parabola, the quadratic formula can also be used to identify the axis of symmetry of the parabola, and the number of real zeros the quadratic equation contains.
If you are trying to solve for x, put in the standard quadratic form: x^2 -x -13 = 0. (ax^2 + bx +c = 0)
Then use the quadratic formula ( https://en.wikipedia.org/wiki/Quadratic_formula )
with x representing an unknown, a, b and c representing constants with a ≠ 0, the quadratic formula is:
{\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\ \ }
where the plus–minus symbol "±" indicates that the quadratic equation has two solutions.[2] Written separately, they become:
{\displaystyle x_{1}={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\quad {\text{and}}\quad x_{2}={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}}
Each of these two solutions is also called a root (or zero) of the quadratic equation. Geometrically, these roots represent the x-values at which any parabola, explicitly given as y = ax2 + bx + c, crosses the x-axis.
As well as being a formula that yields the zeros of any parabola, the quadratic formula can also be used to identify the axis of symmetry of the parabola, and the number of real zeros the quadratic equation contains.
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