Section 5-8

A quadratic equation written in standard form ax 2 + bx + c = 0 can be solved with the Quadratic Formula:

Before using the quadratic formula, we must discover the values of a, b and c from a quadratic equation in standard form: ax 2 + bx + c = 0 Find the values of a, b, and c for the following equation: 3 x 2 +6 x -9 = 0 a = 3 b = 6 c = -9

Remember, the quadratic formula is: To solve the equation above, we simply plug the values of a, b and c into the quadratic formula and evaluate. We already found that a = 3, b = 6 and c = -9. Solve: 3 x 2 +6 x -9 = 0

You Try: Use the Quadratic Formula to solve 3 x 2 – x = 4 (Hint: The equation must be set = 0 to find a, b, and c. ) Remember, the quadratic formula is: Solutions:

There is a connection between the solutions from the Quadratic Formula and the graph of the parabola. You can tell how many x -intercepts (also called the solutions or the roots ) you're going to have from the value inside the square root. The argument of the square root, the expression b 2 – 4 ac, is called the discriminant because, by using its value, you can discriminate between (or tell the differences between) the various solution types: two solutions, one solution, or no solution. Let’s look at an example and compare our solutions to the corresponding graph.

Solve: x ( x – 2) = 4 I cannot apply the Quadratic Formula at this point because the equation is not in standard form. First I need to rearrange the equation in the form ax 2 + bx + c = 0. How do I do that? First, I need to distribute the x on the left-hand side: x ( x – 2) = 4 x 2 – 2 x = 4 Then I'll subtract 4 from both sides ( using the Subtraction Property of Equality ): x 2 – 2 x – 4 = 0 Now I can use the Quadratic Formula!

I need to identify what a, b, and c equal in our equation x 2 – 2 x – 4 = 0. a = 1, b = –2, and c = –4 Now I’m ready to plug into the quadratic formula. If we round to 2 decimal places, are solutions are: x = –1.24 and x = 3.24.

For reference, here's what the graph looks like: There are 2 x -intercepts (roots) and we got 2 solutions when we solved it algebraically.

Solve: 9 x x + 4 = 0 Using a = 9, b = 12, and c = 4, the Quadratic Formula gives: The solution is x = –2 / 3 In the previous examples, I had gotten two solutions because of the "plus-minus" part of the formula. In this case, though, the square root reduced to zero, so the “plus- minus” didn't count for anything. This solution is called a repeated root, because x is equal to –2 / 3, twice: –2 / and –2 / 3 – 0. Any time you get zero in the square root of the Quadratic Formula, you'll only get one solution.

For reference, here's what the graph looks like: There is only 1 x -intercept (root) and we only got 1 solution when we solved it algebraically. The parabola only just touches the x -axis at x = –2 / 3 ; it doesn't actually cross. This is always true: if you have a root that appears exactly twice, then the graph will "kiss" the axis there, but not pass through.

Solve: 3 x x + 2 = 0 Using the Quadratic Formula : a = 3, b = 4, and c = 2 I have a negative number inside the square root. Can I take the square root of a negative? Not that we have learned so far. Therefore our solution is “No Real Numbers”. We have not talked about complex numbers yet but we will in a future unit. I know you are all excited and can’t wait!!

Whether or not you know about complex numbers, you know that you cannot graph your solution because you cannot graph the square root of a negative number since there are no such values on the x -axis. Since you can't find a “graphable” solution to the quadratic, then reasonably there should not be any x -intercepts. Here's the graph: This relationship is always true: If you get a negative value inside the square root, then there will be no real number solution, and therefore no x -intercepts.