Algebra 2

The Quadratic Formula: Solutions and the Discriminant

Solve x (x – 2) = 4. Round your answer to two decimal places.

I not only cannot apply the Quadratic Formula at this point, I cannot factor either. Why? Because this equation is yet in the correct form.

And I certainly can not claim, with a straight face, that "x = 4, x – 2 = 4", because this is not how "solving by factoring" works.

No matter which solution method I intend to use — whether I'm factoring or using the Quadratic Formula to find my answers — I must first rearrange the equation into the form "(quadratic) = 0".

MathHelp.com

The Quadratic Formula

The first thing I'll do here is multiply through on the left-hand side, and then I'll move the 4 over from the right-hand side to the left-hand side:

x (x – 2) = 4

x2 – 2x = 4

x2 – 2x – 4 = 0

Since there are no factors of (1)(–4) = –4 that add up to –2, then this quadratic does not factor. (In other words, there is no possible way that the faux-factoring solution of "x = 4, x – 2 = 4" could be even slightly correct.)

So factoring won't work, but I can use the Quadratic Formula; in this case, I'll be plugging in the values a= 1, b = –2, and c = –4:

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

= \dfrac{2 \pm \sqrt{4 + 16\,}}{2} = \dfrac{2 \pm \sqrt{20\,}}{2}

= \dfrac{2 \pm \sqrt{4\,}\sqrt{5\,}}{2} = \dfrac{2 \pm 2\, \sqrt{5\,}}{2}

= \dfrac{2\,\left(1 \pm \sqrt{5\,}\right)}{2(1)} = 1 \pm \sqrt{5\,}

\approx 1.236068,\, 3.236968

Then the answer is:

x = –1.24, x = 3.24, rounded to two decimal places.

For reference, here's what the graph of the associated quadratic, y = x2 – 2x – 4, looks like:

As you can see, the solutions from the Quadratic Formula match up with the x-intercepts. The locations where the graph crosses the x-axis give the values that solve the original equation.

There is another connection between the solutions from the Quadratic Formula and the graph of the parabola: you can tell how many x-intercepts you're going to have from the value inside the square root. The argument (that is, the contents) of the square root, being the expression b2 – 4ac, is called the "discriminant" because, by using its value, you can "discriminate" between (that is, be able to tell the difference between) the various solution types.

In this case, the value of the discriminant b2 – 4ac was 20; in particular, the value was not zero and was not negative. Because the value was not negative, the equation was going to have at least one (real-valued) solution; because the value was not zero, the two solutions were going to be distinct (that is, they were going to be different from each other).

Solve 9x2 + 12x + 4 = 0. Leave your answer in exact form.

Using a = 9, b = 12, and c = 4, the Quadratic Formula gives me:

x = \dfrac{-(12) \pm \sqrt{(12)^2 - 4(9)(4)\,}}{2(9)}

= \dfrac{-12 \pm \sqrt{144 - 144\,}}{18}

= \dfrac{-12 \pm \sqrt{0\,}}{18} = \dfrac{-12 \pm 0}{18}

= \dfrac{-12}{18} = \dfrac{-2}{3} = -\dfrac{2}{3}

Then the answer is:

\mathbf{\color{purple}{\mathit{x} = -\dfrac{2}{3}}}

In the first example on this page, I had gotten two solutions because the value of the discriminant (that is, the value inside the square root) was non-zero and positive. As a result, the "plus-minus" part of the Formula gave me two distinct values; one for the "plus" part of the numerator and another for the "minus" part. In this case, though, the square root reduced to zero, so the plus-minus didn't count for anything.

This sort of solution, where you get only one value because "plus or minus zero" didn't change anything, is called a "repeated" root, because x is equal to

- \frac{2}{3}

, but it's equal to this value kind-of twice:

- \frac{2}{3} + 0

and

- \frac{2}{3} - 0

http://www.purplemath.com/modules/quadform2.htm

Algebra 1

SIMPLIFYING RADICALS

To simplify radicals, rather than looking for perfect squares or perfect cubes within a number or a variable the way it is shown in most books, I choose to do the problems a different way, and here is how.

Here are the steps required for Simplifying Radicals:

Step 1:	Find the prime factorization of the number inside the radical. Start by dividing the number by the first prime number 2 and continue dividing by 2 until you get a decimal or remainder. Then divide by 3, 5, 7, etc. until the only numbers left are prime numbers. Click on the link to see some examples of Prime Factorization. Also factor any variables inside the radical.
Step 2:	Determine the index of the radical. The index tells you how many of a kind you need to put together to be able to move that number or variable from inside the radical to outside the radical. For example, if the index is 2 (a square root), then you need two of a kind to move from inside the radical to outside the radical. If the index is 3 (a cube root), then you need three of a kind to move from inside the radical to outside the radical.
Step 3:	Move each group of numbers or variables from inside the radical to outside the radical. If there are nor enough numbers or variables to make a group of two, three, or whatever is needed, then leave those numbers or variables inside the radical. Notice that each group of numbers or variables gets written once when they move outside the radical because they are now one group.
Step 4:	Simplify the expressions both inside and outside the radical by multiplying. Multiply all numbers and variables inside the radical together. Multiply all numbers and variables outside the radical together.

Example 1 – Simplify:

Step 1: Find the prime factorization of the number inside the radical.
Step 2: Determine the index of the radical. In this case, the index is two because it is a square root, which means we need two of a kind.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the pair of 2’s and 3’s moved outside the radical.
Step 4: Simplify the expressions both inside and outside the radical by multiplying.

Example 2 – Simplify:

Step 1: Find the prime factorization of the number inside the radical and factor each variable inside the radical.
Step 2: Determine the index of the radical. In this case, the index is three because it is a cube root, which means we need three of a kind.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the 3’s, x’s (two groups), and y’s moved outside the radical.
Step 4: Simplify the expressions both inside and outside the radical by multiplying.

Click Here for Practice Problems
Example 3 – Simplify:

Step 1: Find the prime factorization of the number inside the radical and factor each variable inside the radical.
Step 2: Determine the index of the radical. In this case, the index is five because it is a fifth root, which means we need five of a kind.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the 3’s, x’s, and y’s moved outside the radical.
Step 4: Simplify the expressions both inside and outside the radical by multiplying.

Click Here for Practice Problems

Example 4 – Simplify:

Step 1: Find the prime factorization of the number inside the radical and factor each variable inside the radical.
Step 2: Determine the index of the radical. In this case, the index is two because it is a square root, which means we need two of a kind.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the 2’s, 3’s, x’s (two groups), and y’s (four groups) moved outside the radical.
Step 4: Simplify the expressions both inside and outside the radical by multiplying.

Click Here for Practice Problems
Example 5 – Simplify:

Step 1: Find the prime factorization of the number inside the radical and factor each variable inside the radical.
Step 2: Determine the index of the radical. In this case, the index is four because it is a fourth root, which means we need four of a kind.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the pair of 2’s and y ’s moved outside the radical.
Step 4: Simplify the expressions both inside and outside the radical by multiplying.

Click Here for Practice Problems

Example 6 – Simplify:

Step 1: Find the prime factorization of the number inside the radical and factor each variable inside the radical.
Step 2: Determine the index of the radical. In this case, the index is three because it is a cube root, which means we need three of a kind.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the 2’s, 3’s, x’s, and y’s (two groups) moved outside the radical.
Step 4: Simplify the expressions both inside and outside the radical by multiplying.