Slide 1Slide 2Slide 3Slide 4The Quadratic FormulaSlide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20The Discriminant and the Kinds of Solutions to ax2 + bx +c = 0Slide 22Slide 23Slide 24Slide 25Slide 26REMINDER!!!Reminder:Slide 29Any questions on the Section 5.8 homework?Pass your worksheets for this assignment to the middle aisle for pickup now.Remember the problem like this one from the homework that was due today?Wouldn’t it be nice if there was an easier way to do it than by factoring?Leave factoring up on board: (4x - 9)(3x + 8)PleaseCLOSE YOUR LAPTOPSand turn off and put away your cell phones.Sample Problems Page Link(Dr. Bruce Johnston)Section 8.2The Quadratic FormulaThe Quadratic FormulaThe quadratic formula is another technique we can use for solving quadratic equations.Remember, quadratic equations are polynomial equations of degree 2, such as x2 + 3x -7 = 0 or 5x2 – 14 = 0.The quadratic formula is derived from a process called “completing the square” for a general quadratic equation. –See Section 8.1 if you’re interested in seeing how this formula is derived. –This will also be covered in Math 120 in more detail, along with the technique called “completing the square”.The Quadratic Formula: aacbbx242The solutions to the equation ax2 + bx + c = 0 are given by the formulaNote: This formula IS on the pink formula sheet, but you’ll probably have it memorized by the time you’ve done the first few homework problems.The Big Question: How can we tell when we should use factoring and when we should use the quadratic formula?Solve x2 + 4x + 3 = 0 by •Factoring•The quadratic formula.Which way works best?Example 1Solve x2 + 4x + 3 = 0 by Factoring: This one is pretty easy to factor. The factoring is (x + 3)(x + 1) = 0, so the solutions are given by x + 3 = 0, or x = -3, and x + 1 = 0, which gives x = -1.Now, solve x2 + 4x + 3 = 0 by the quadratic formula: a = 1, b = 4, c = 3, so the formula gives: 32622412222422424421216412314442orxWhich way works best for this problem? In this case, the factoring method is much quicker, although BOTH methods give the same answer.Solve x2 + 5x + 12 = 0 by •Factoring•The quadratic formula.Which way works best?Example 2Solve x2 + 5x + 12 = 0 by Factoring: This one looks pretty easy to factor, but when you start trying to find two factors of 12 that add up to 5, nothing works. (1+12=13, 2+6=8, 3+4=7). What does this mean? It means that the polynomial is PRIME, and there are no rational solutions. (Remember, a rational number is either an integer or a fraction.)Solve x2 + 5x + 12 = 0 (continued):•Let’s see what the quadratic formula gives in this case: a = 1, b = 5, c = 12so the formula gives: 2235248255121214552xNotice that the number under the radical sign is negative, which means there are no real answers. If the number under the square root sign comes out to be positive but it’s not a perfect square, this means the answer is a real number, but is irrational because it can’t be simplified to remove the radical. In either of these cases, we’d say the polynomial is prime, and therefore has no rational roots.So which way works best for solving x2 + 5x + 12 = 0?Either way works fine, but if you think a polynomial is prime, a good way to check is by calculating the discriminant (b2 – 4ac). If the discriminant is either negative or not a perfect square, then you know for sure that your polynomial is prime and there are no rational solutions.Now re-do this problem from the 5.8 homework using the quadratic formula:Answers: -8/3, 9/4Which way works best in this case?Either way works, but the quadratic formula approach is probably going to be faster than factoring for most people.Moral of the story: For a quadratic equation with a leading coefficient other than 1, it’s probably going to be quicker to solve it using the quadratic formula than it would be to factor the polynomial.)1(2)20)(1(4)8(82x280648214482128210,24220 or or  x2 + 8x – 20 = 0 (multiply both sides by 8) a = 1, b = 8, c = -208125Solve x2 + x – = 0 by the quadratic formula.Question: What if some coefficients in your quadratic equation are fractions? ANSWER: Clear them first by multiplying all terms by the LCD:•The expression under the radical sign in the quadratic formula (b2 – 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 (if it’s positive), one real solution (if it’s zero), or two complex, but not real solutions (if it’s negative – a topic to be discussed in Math 120).No x-interceptsNo real solution; two complex imaginary solutionsb2 – 4ac < 0One x-interceptOne real solution (a repeated solution)(If b2 – 4ac is a perfect square, the solution will be a rational number. If not, it’s irrational.)b2 – 4ac = 0Two x-interceptsTwo unequal real solutions(If b2 – 4ac is a perfect square, the two solutions will be rational numbers. If not, they’re both irrational.)b2 – 4ac > 0Graph of y = ax2 + bx + cKinds of solutions to ax2 + bx + c = 0Discriminantb2 – 4acThe Discriminant and the Kinds of Solutions to ax2 + bx +c = 0Use the discriminant to determine the number and type of solutions for the following equation.5 – 4x + 12x2 = 0 a = 12, b = -4, and c = 5b2 – 4ac = (-4)2 – 4(12)(5) = 16 – 240 = -224 Since the discriminant is negative, there are no real solutions. Question: What would this graph look like?ExampleUse the discriminant to determine the number and type of solutions for the following equation.25x2 - 4 = 0 a = 25, b = 0 (why?) , and c = -4 b2 – 4ac = (0)2 – 4(25)(-4) = 0 – -400 = 400Since the discriminant is positive, there are two real solutions. (You could go on to show that the solutions are 2/5 and -2/5, either by factoring or using the quadratic formula.)ExampleUse the discriminant to determine the number and type of solutions for the following equation.5 – 4x + 12x2 = 0 a = 12, b = -4, and c = 5 b2 – 4ac = (-4)2 – 4(12)(5) = 16 – 24 = -224Since the discriminant is negative, there are no real solutions.