Week of April 17,2017

Algebra1

The Quadratic Formula

Often, the simplest way to solve "ax2 + bx + c = 0" for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it doesn't factor at all, or you just don't feel like factoring. While factoring may not always be successful, the Quadratic Formula can always find the solution.

The Quadratic Formula uses the "a", "b", and "c" from "ax2+ bx + c", where "a", "b", and "c" are just numbers; they are the "numerical coefficients" of the quadratic equation they've given you to solve.

The Quadratic Formula is derived from the process of completing the square, and is formally stated as:

The Quadratic Formula: For ax2 + bx + c = 0, the values of x which are the solutions of the equation are given by:

$x = \dfrac{-b \pm\sqrt{b^2 - 4ac\,}}{2a}$

For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic) = 0". Also, the "2a" in the denominator of the Formula is underneath everything above, not just the square root. And it's a "2a" under there, not just a plain "2". Make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back in" on your test, and you'll mess yourself up. Remember that "b2" means "the square of ALL of b, including its sign", so don't leave b2 being negative, even if b is negative, because the square of a negative is a positive.

In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run. Trust me on this!

Here are some examples of how the Quadratic Formula works:

• Solve x2 + 3x – 4 = 0

This quadratic happens to factor:

x2 + 3x – 4 = (x + 4)(x – 1) = 0

...so I already know that the solutions are x = –4 and x = 1. How would my solution look in the Quadratic Formula? Using a = 1, b = 3, and c = –4, my solution looks like this:

$x = \dfrac{-(3) \pm \sqrt{(3)^2 - 4(1)(-4)\,}}{2(1)}$

$= \dfrac{-3 \pm \sqrt{9 + 16\,}}{2} = \dfrac{-3 \pm \sqrt{25\,}}{2}$

$= \dfrac{-3 \pm 5}{2} = \dfrac{-3 - 5}{2},\, \dfrac{-3 + 5}{2}$

$= \dfrac{-8}{2},\, \dfrac{2}{2} = -4,\, 1$

Then, as expected, the solution is x = –4, x = 1.

Algebra 2

Assessments: Unit 4 test is Tuesday

Monday
We are reviewing rational expressions  and equations for a test.
Help:https://www.khanacademy.org/math/algebra2/rational-expressions-equations-and-functions
HW; Review sheet

Tuesday
Test day, you shall shine!
Help:http://www.purplemath.com/modules/solvrtnl.htm
HW; None

Wednesday
 You can wrap up the test today, write on rational expressions and we look at basic radical equations
Help:https://www.khanacademy.org/math/algebra2/radical-equations-and-functions/solving-square-root-equations/v/solving-radical-equations
HW; None

Thursday
It is day two of radical equations!
Help:http://www.mathplanet.com/education/algebra-1/radical-expressions/radical-equations
HW; None

Friday
Yes, it is Friday!
Day three of radical equations with radicals on both sides.
Help:https://www.mathsisfun.com/algebra/radical-equations-solving.html
HW; None


Have a great weekend!