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Please CLOSE YOUR LAPTOPS, and turn off and put away your cell phones, and get out your note-taking materials. Today’s daily quiz will be given at the end of the hour.

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Section 11.2 The Quadratic Formula

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Remember the problem like this one you had to solve by factoring in the homework back in Chapter 6? Wouldn’t it be nice if there was an easier way to do this problem than by factoring? Since there’s no GCF that we can take out, we can either “guess and check” (which is harder than usual because there are so many possible combinations to try), or we can use the British method. Even that method is a challenge, since it starts with multiplying 12*(-72) = -864, then trying to find two factors of -864 that add up to 5. The factors that work are 32 and -27, so you’d have 12x 2 +32x – 27x -72 = 0. Factoring by grouping then gives the correct factoring: (4x – 9)(3x + 8) = 0, which leads to the two solutions of 9/4 and -8/3.

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The Quadratic Formula The quadratic formula is another technique we can use for solving quadratic equations. Remember, quadratic equations are polynomial equations of degree 2, such as x 2 + 3x -7 = 0 or 5x 2 – 14 = 0.

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The quadratic formula is derived from a process called “completing the square” for a general quadratic equation. – See Section 11.1 in the online textbook if you’re interested in seeing how this formula is derived. – If you are taking Math 120 next, this will be covered in that class in more detail, along with the technique called “completing the square”.

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The Quadratic Formula: The two solutions to the equation ax 2 + bx + c = 0 (where a is the coefficient of the x 2 term, b for the x term, and c is the constant term.) are given by the formula Notes: 1.The symbol ± means “plus or minus”. 2.Using the + version gives you one solution, and using the ˗ version gives you the second solution. 3.This formula IS on the formula sheet, but you’ll probably have it memorized by the time you’ve done the first few homework problems.

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Solve x 2 + 4x + 3 = 0 by Factoring The quadratic formula. Which way works best? Example 1

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Solve x 2 + 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, giving x = -3, and x + 1 = 0, giving x = -1.

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Now, solve x 2 + 4x + 3 = 0 by the quadratic formula: a = 1, b = 4, c = 3, so the formula gives:

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Which way works best for this problem? In this case, the factoring method is much quicker, although BOTH methods give the same answer. And by the way, how would we check these answers on a quiz or a test?

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Back to this problem: Answers: -8/3, 9/4 Write the steps for this problem in your notebook while I work it on the board. (We’ll probably want to use a calculator to do the multiplications in this problem.) We’ve seen that factoring using the British method is a lot of work. Let’s try using the quadratic formula instead.

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Solve x 2 + 5x + 12 = 0 by Factoring The quadratic formula. Which way works best? Example:

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Solve x 2 + 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.)

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If you think a polynomial is prime, a good way to check is by calculating the discriminant (the b 2 – 4ac part under the radical in the quadratic formula). If the discriminant is negative, then you know for sure that your polynomial is prime and there are no real solutions. (There will be two complex but not real solutions involving the square root of -1, which is given the symbol “ i ”.) If the discriminant is positive and a perfect square, then you will get two real solutions that are real and rational numbers. If the discriminant is positive but not a perfect square, then you know for sure that your polynomial is prime (can’t be factored without using radicals), but there will be two irrational solutions that will contain radicals. If the discriminant is zero, there is only one real solution, and the two factors of the quadratic trinomial will be the same binomial.

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Solve x 2 + 5x + 12 = 0 (continued): Let’s see what the quadratic formula gives in this case: a = 1, b = 5, c = 12 so the formula gives: Notice that the number under the radical sign is negative, which means there are no real answers. There will be two complex but not real solutions involving the symbol “ i ” which is the square root of -1.

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Solve x 2 + 2x - 4 = 0 by Factoring The quadratic formula. Which way works best? Example:

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Solve x 2 + 2x - 4 = 0 by factoring: Again, this one looks like it would be easy to factor, but when you start trying to find two factors of 4 that add up to 2, nothing works. (-1 + 4 = 3, -4 + 1=-3, -2+2 = 0) What does this mean? It means that the polynomial is PRIME, and there are no rational solutions. (Remember, a rational number is a real number that is either an integer or a fraction.)

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Solve x 2 + 2x - 4 = 0 (continued): Let’s see what the quadratic formula gives in this case: a = 1, b = 2, c = -4 so the formula gives: Notice that the number (20) under the radical sign is positive, which means there are two real answers. However, since 20 is not a perfect square, its square root is irrational. In this case, we’d again say that the polynomial is prime, because it has no rational roots.

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Solve x 2 + 2x - 4 = 0 using the quadratic formula (continued): IMPORTANT NOTE: Because the discriminant (20) is POSITIVE, we still do have two REAL NUMBER solutions, but they won’t be rational numbers because 20 is not a perfect square. If we are asked to find EXACT answers, that means we have to use radicals in the answer, but first we need to check to see if the radical can be simplified.

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Solve x 2 + 2x - 4 = 0 using the quadratic formula (continued):

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Step 1: Multiply every term by 8 to get x 2 + 8x – 20 = 0 Therefore a = 1, b = 8, c = -20 Solve x 2 + 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: + version gives: 4/2 = 2 - version gives -20/2 = -10 IMPORTANT: When you check these answers, plug them back into the ORIGINAL equation, the one with the fractions, just in case you made a mistake when you cleared the fractions in Step 1.

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The expression under the radical sign in the quadratic 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 (if it’s positive), one real solution (if it’s zero), or two complex but not real solutions (if it’s negative; this type is discussed further in Math 120).

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No x-intercepts No real solution; two complex imaginary solutions b 2 – 4ac < 0 One x-intercept One real solution (a repeated solution) (If b 2 – 4ac is a perfect square, the solution will be a rational number. If not, it’s irrational.) b 2 – 4ac = 0 Two x-intercepts Two unequal real solutions (If b 2 – 4ac is a perfect square, the two solutions will be rational numbers. If not, they’re both irrational.) b 2 – 4ac > 0 Graph of y = ax 2 + bx + c Kinds of solutions to ax 2 + bx + c = 0 Discriminant b 2 – 4ac The discriminant and the kinds of solutions to ax 2 + bx +c = 0

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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 Since the discriminant is negative, there are no real solutions. (The two non-real solutions will be complex numbers involving the symbol “ i ”.) Question: What would this graph look like? Example

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Use the discriminant to determine the number and type of solutions for the following equation. 25x 2 - 4 = 0 a = 25, b = 0 (why?), and c = -4 b 2 – 4ac = (0) 2 – 4(25)(-4) = 0 – -400 = 400 Since 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.) Example

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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 – 24 = -224 Since the discriminant is negative, there are no real solutions. NOTE: There will be two COMPLEX solutions that are not real numbers, i.e. they will contain the imaginary number i which is the square root of -1. Example

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Use the discriminant to determine the number and type of solutions for the following equation. x 2 – 8x + 16 = 0 a = 1, b = -8, and c = 16 b 2 – 4ac = (-8) 2 – 4(1)(16) = 64 – 64 = 0 Since the discriminant is zero, there is one real solution. (You could go on to show that the solution is 4, either by factoring or using the quadratic formula.) Question: What would this graph look like? Example

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REMINDER: The assignment on today’s material (HW 42) is due at the start of the next class session. Lab hours in 203: Mondays through Thursdays 8:00 a.m. to 7:30 p.m. Please remember to sign in on the Math 110 clipboard by the front door of the lab

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Please open your laptops and find Daily Quiz 36. A scientific calculator may be used on this quiz. You may also use your yellow formula sheet. Remember to turn in your answer sheet to the TA when the quiz time is up.

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Please open your laptops, log in to the MyMathLab course web site, and open Quiz 5.3/4. You may use the formula sheet on this quiz – please don’t write.

Please open your laptops, log in to the MyMathLab course web site, and open Quiz 5.3/4. You may use the formula sheet on this quiz – please don’t write.

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