I hate MATH!

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Tanis143

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So whatever happened to FOIL (First - Outside - Inside - Last), or am I just hopelessly outdated?
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.
 

John6185

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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.
 

Tanis143

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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.

A lot of people have/had problems with math because so few teachers explain it, they just have you memorize tables and go from there. Its like the difference between showing a person how to build an engine and then expecting them to be able to troubleshoot vs explaining how an engine works first, then showing how to build it. Once you understand how math works, the rest falls into place.
 

Rod Snell

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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.
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}}\ \ }
{\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}}}
{\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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Snattlerake

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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}}\ \ }
{\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}}}
{\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.
See why I hate math?
 

Tanis143

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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}}\ \ }
{\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}}}
{\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.

In this case, the + version gives the answer that gets the closest due to 52 not having a solid square root. The answer would be 4.14 (comes out to only off by .0004. Its actual answer would be 𝑥=1+√53/2.
 

Cowbaby

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If I remember right there are 3 ways to solve an equation in quadratic form ax^2+bx+c. You can try to factor something out and make it simpler and then solve which can only be done sometimes, the aforementioned FOIL method and the equation method which always works just plugging in values. So after that I always found the other two a waste of time.
I remember thinking that was the end all hard stuff when I took algebra and cussing quadratic equations and then you get to calculus years and end up praying for them because the solution is always in the form of a nice simple Ax+b and didn't take half a page to get to something you could solve. You got so good at doing that stuff you did the algebra and trig part in your head believe it or not.

But man I have slept since them and that was 40 years ago. I guess that is a testament to those old battleaxe math professors I had that I can even recall anything at all. That was back in the day where we didn't get participation trophies and it was 3 tests and a comprehensive final ALWAYS. You had to get it and keep up or get left behind.

I personally always found statistics the most useful in my life for what I needed of the math I was exposed to. Even that stuff didn't make sense until one day an old Professor in a non math class who retired from Firestone took a bunch of data, made a dot plot that was trending upwards and to the left on a x, y axis. He slapped a line down the center turned the paper sideways and drew a bellcurve over the middle and it was like a light bulb finally went of in my head. Hey, this stuff is bad ass. Then I got real interested.
 
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