Question

Can someone explain to me how to solve quadratic problems?

Answer 1

1Combine all of the like terms and move them to one side of the equation. The first step to factoring an equation is to move all of the terms to one side of the equation, keeping the x2 term positive. To combine the terms, add or subtract all of the x2 terms, the x terms, and the constants (integer terms), moving them to one side of the equation so that nothing remains on the other side. Once the other side has no remaining terms, you can just write "0" on that side of the equal sign. Here's how you do it:[1]
2x2 - 8x - 4 = 3x - x2 =
2x2 +x2 - 8x -3x - 4 = 0
3x2 - 11x - 4 = 0

2Factor the expression. To factor the expression, you have to use the factors of the x2 term (3), and the factors of the constant term (-4), to make them multiply and then add up to the middle term, (-11). Here's how you do it:
Since 3x2 only has one set of possible factors, 3x and x, you can write those in the parenthesis: (3x +/- ? )(x +/- ?) = 0.
Then, use process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4.
Try out (3x +1)(x -4). When you multiply them out, you get - 3x2 -12x +x -4. If you combine the terms -12x and x, you get -11x, which is the middle term you were aiming for. You have just factored the quadratic equation.
As an example, let's try a factoring combination that does not work: (3x -2)(x +2) = 3x2 +6x -2x -4. If you combine those terms, you get 3x2 -4x -4. Though the factors -2 and 2 do multiply to make -4, the middle term does not work, because you wanted to get -11x, not -4x.

3 Set each set of parenthesis equal to zero as separate equations. This will lead you to find two values for x that will make the entire equation equal to zero. Now that you've factored the equation, all you have to do is put each set of parenthesis equal to zero. So, you should write 3x +1 = 0 and x - 4 = 0.

If this isn't enough, please let me know.

Answer 2

well it depends on the type of quadratic but the fail safe method is using the general quadratic formula

for a quadratic

\[ax^2 + bx + c = 0\]

the formula is

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

so for a problem such as

\[x^2 + 5x + 6 = 0\]

a = 1, b = 5 and c = 6

substitute them into the formula and calculate.

you'll find

x = -2 and x = -3

here are some good notes that explain solving quadratics and the different methods you can use

http://tutorial.math.lamar.edu/Classes/Alg/SolveQuadraticEqnsI.aspx

Answer 3

Oh kay thanks!

Answer 4

So did that help at all @HailKK? If not, please let me know. I know it was long but it should be helpful.

Answer 5

@Days.Of.Future.Past

Its an interesting set of notes you posted. Did you also write these notes ?

http://www.wikihow.com/Solve-Quadratic-Equations

if not, please do the right thing by the author and acknowledge the source.