1. Solving Quadratic Equations Unit Review

2. Solving Quadratics By Graphing

3. Graphs of Quadratic Equations The graph of a quadratic equation is a parabola. The highest point or lowest point on the parabola is the vertex. Axis of symmetry is the line that runs through the vertex and through the middle of the parabola.

4. Graph by plotting points.

5. Intercepts of a Porabola Although we can simply plot points, it is helpful to know some information about the parabola we will be graphing prior to finding individual points. To find x-intercepts of the parabola, let y = 0 and solve for x. To find y-intercepts of the parabola, let x = 0 and solve for y.

6. Characteristics of the Porabola If the quadratic equation is written in standard form, y = ax 2 + bx + c, 1) the parabola opens up when a > 0 and opens down when a < 0. 2) the x-coordinate of the vertex is To find the corresponding y-coordinate, you substitute the x-coordinate into the equation and evaluate for y.

7. Solve By Graphing:

8. Graph:

9. Solution Types: a) 2 Solutions b) No Solution c) 1 Solution

10. Solving Quadratics By Factoring

11. Solving by Factoring Equation Needs to be in Standard Form! Zero Product Property: For any real numbers a and b, if ab=0, then either a=0, b=0, or both.

12. Remember Factoring! Factoring when a = 1 Factoring when 0 > a > 0 To solve by factoring Get equations into standard form Factor the quadratic using what we learned in the previous chapter. Set each factored term equal to zero. Solve for x in each factored term. Check your solutions!!

13. Solve by factoring x 2 + 3x = 28 x 2 + 3x – 28 = 0 (x + 7) (x – 4) = 0 x + 7 = 0 and x – 4 = 0 x = -7 x = 4 Solutions: x = -7 and 4 Check ✓

14. Example 2x 2 + 4x – 30 = 0 (x + 5) (2x - 6) = 0 x + 5 = 0and2x – 6 = 0 x = -5x = 3 Solutions: x = -5 and 3 Check ✓

15. Solving Quadratics by Quadratic Formula

16. The Quadratic Formula Another technique for solving quadratic equations is to use the quadratic formula. The formula is derived from completing the square of a general quadratic equation.

17. A quadratic equation written in standard form, ax 2 + bx + c = 0, has the solutions.

18. Using the Quadratic Formula Solve 11n 2 – 9n = 1 by the quadratic formula. 11n 2 – 9n – 1 = 0, so a = 11, b = -9, c = -1

19. Solve (1/8) x 2 + x – (5/2) = 0 by the quadratic formula. x 2 + 8x – 20 = 0 (multiply both sides by 8) a = 1, b = 8, c =  20

20. Solve x(x + 6) =  30 by the quadratic formula. x 2 + 6x + 30 = 0 a = 1, b = 6, c = 30 So there is no real solution.

21. The Discriminant: The expression under the radical sign in the formula (b 2 – 4ac) is called the discriminant. The discriminant will take on a value that is positive, 0, or negative. The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively.

22. Use the discriminant to determine the number and type of solutions for the following equation. 5 – 4x + 12x 2 = 0 a = 12, b = –4, and c = 5 b 2 – 4ac = (–4) 2 – 4(12)(5) = 16 – 240 = –224 There are no real solutions.

23. When the discriminant = 0 there is one solution When the discriminant > 0 there is 2 solutions When the discriminant < 0 there are no solutions

24. Solving Quadratic Equations Steps in Solving Quadratic Equations 1)If the equation is in the form (ax+b) 2 = c, use the square root property to solve. 2)If not solved in step 1, write the equation in standard form. 3)Try to solve by factoring. 4)If you haven’t solved it yet, use the quadratic formula.

25. Word Problem Application! The diagram below shows a pattern of an open-top box. The total area of the sheet is 288 inches square. The height of the box is 3 in. Therefore, 3- in. by 3-in. squares are cut from each corner. Find the dimensions of the box.

26. Word Problem Application! An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t 2 + 19.6t + 58.8, where s is in meters. When does the object strike the ground? 0 = –4.9t 2 + 19.6t + 58.8 0 = t 2 – 4t – 12 0 = (t – 6)(t + 2) Then t = 6 or t = –2. The second solution is from two seconds before launch, which doesn't make sense in this context. (It makes sense on the graph, because the line crosses the x-axis at –2, but negative time won't work in this word problem.) So "t = –2" is an extraneous solution, and I'll ignore it.