Solving Quadratic Equations Using the QUADRATIC FORMULA

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A quadratic equation always equals zero, and always has a squared term. A properly arranged quadratic equation is modeled like this:

ax² + bx +c = 0

'a', 'b' and 'c' are all coefficients (numbers) and they may be positive OR negative. Here is an example:

4x² - 17x -15 = 0

HOW TO SOLVE A QUADRATIC EQUATION

There are many ways to solve a quadratic equation. The three main ways are completing the square, using the Quadratic Formula and factoring. This lesson demonstrates how to use the Quadratic Formula to solve a quadratic equation.

USING THE QUADRATIC FORMULA
The Quadratic Formula may seem intimidating, but it is really just a system you use to “plug in” the coefficient values for the variables ‘a’, ‘b’ and ‘c’, just like any other formula. It is important that you learn it well, because out of all three popular methods of solving quadratic equations, the Quadratic Formula is the only one that works every time. Some of the steps in the formula, such as radicals, may mean that you have to elevate your game.
One way to learn how to memorize the Quadratic Formula is by learning a common song that sings the Quadratic Formula to a familiar tune. There are a variety of videos available. HERE IS ONE.

Here is the actual formula:

When I say it out loud, it sounds like this:
“Negative ‘b’ plus or minus the square root of ‘b’ squared, minus four ‘a’ ‘c’, all over two ‘a’ .”
When we plug in the actual coefficient numbers for the variables in the formula, it looks like this:

We start by dealing with the first ‘b’ term. By having a negative in front of the entire coefficient, it is basically saying that we simply change the sign. Now our formula looks like this:

Now we square our next ‘b’ term. The answer will always be a positive number, since a negative number times a negative number equals a positive number. Our changed formula:

At this point, it is time to multiply the ‘a’ and ‘c’ terms, which leaves us with this:

Next, multiply the result of the previous step by (-4):

Add the two numbers left in the discriminant:

Now we find out what the square root of 529 is. I should mention that, at any point in the process, you may solve the bottom terms. Our formula is beginning to look like some easy math by now:

What this means is that we will wind up with two separate answers, as expected. The first answer will be 17 plus 23, over 8, and the other answer will be 17 minus 23, over 8:

solution is x = 5 and -3/4. This means that if you put either number into our original equation, you will make it equal zero. These two points are also where the graphed parabola crosses the x-axis. These are also known as ‘roots’. The ‘solutions’ are the actual ‘x’ values for the coordinates that cross the x-axis: (5, 0) and (-3/4, 0)

Steve Wilson, a.k.a. "Steve the Tutor," is a tutor in the Tri-Cities (Kennewick, Pasco, Richland), WA, and surrounding areas. With a wide range of experience and adaptability, you will find that Steve the Tutor is the tutor for you! For more information, please visit my Wyzant profile. Thanks!

Keywords:algebra,quadratic equation,factoring,tutor,steve,washington,pasco,kennewick,richland,wyzant

Solving quadratic equations by FACTORING

http://www.wyzant.com/tutors/tutorsteve

A quadratic equation always equals zero, and always has a squared term. A properly arranged quadratic equation is modeled like this:

ax² + bx +c = 0

'a', 'b' and 'c' are all coefficients (numbers) and they may be positive OR negative. Here is an example:

4x² - 17x -15 = 0

HOW TO SOLVE A QUADRATIC EQUATION

There are many ways to solve a quadratic equation. The three main ways are completing the square, using the Quadratic Formula and factoring. This lesson demonstrates how to use factoring to solve a quadratic equation.

FACTORING

Multiply the 'a' coefficient by the 'c' coefficient. Ignore signs for now.

4 ('a' coefficient) x 15 ('c' coefficient) = 60

Find the factors of 60 ('what' times 'what' equals 60), looking for a pair that equal -17 (the 'b' coefficient) when added together:

Factors of 60: [1, 60] : [2, 30] : [3, 20] : [4, 15] : [5, 12] : [6, 10] 

Out of all of these pairs, only one pair works, and that is only if you have the signs in a specific order:

-20 + 3 = -17

Now you set up two fresh binomials with our 'a' and 'c' terms, which look like this:

(4x² +      ) and (      - 15)

You then take the two numbers that you factored, and put them, combined with an 'x', in the missing places in the binomial model. It doesn't matter which one you put where.

(4x² - 20x) and (3x -15)

Now you factor out the common terms, also known as "Reverse Distribution":

4x(x-5) and 3(x-5)

If the new terms in the binomials match EXACTLY, then that is one of your solution sets. In our example, both sets now contain exactly (x-5). Combine the factors on the outside of the binomials. This is your second solution set:

4x(x-5) and 3(x-5)

(x-5) (4x+3)

Use the zero property to solve for 'x' for both sets:

x-5 = 0

x = 5

4x+3 = 0

4x = -3

x = -3/4

Your solution is x = 5 and -3/4. This means that if you put either number into our original equation, you will make it equal zero. These two points are also where the graphed parabola crosses the x-axis. These are also known as ‘roots’. The ‘solutions’ are the actual ‘x’ values for the coordinates that cross the x-axis: (5, 0) and (-3/4, 0)

Steve Wilson, a.k.a. "Steve the Tutor," is a tutor in the Tri-Cities (Kennewick, Pasco, Richland), WA, and surrounding areas. With a wide range of experience and adaptability, you will find that Steve the Tutor is the tutor for you! For more information, please visit my Wyzant profile. Thanks!

http://www.wyzant.com/tutors/tutorsteve

Keywords:algebra,quadratic equation,factoring,tutor,steve,washington,pasco,kennewick,richland,wyzant