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M$5: Intermediate Algebra & Trigonometry Mr. Tamhane 2009-2010 Name: _______________________ Date: _______________________ Sum and Product of the Roots of Quadratic Equations Worksheet 1. One root of x 2 kx 12 0 is 3. Find k. Product = 12, so 3(x) = 12 x=4 3 +4 = 7, but I’ll use -7 in the equation. So, k=-7 2. The roots of the equation x 2 4 x c 0 are 2+3i and 2-3i. What is the value of c? C is the product of the two roots. (2+3i)(2-3i)=4-6i+6i-9i2 = 4 + 9 = 13 3. A quadratic equation having the roots 0 and -2 is: Sum: -2, but I’ll use 2 in the equation Product: 0 So my answer is x2+2x+0=0 or more simply x2+2x=0 4. Write, in simplest form, a quadratic equation whose roots are k and –k. Sum: 0 Product: -k2 So x2+0x-k2=0 or more simply, x2-k2=0 5. Given the equation x 2 kx 7 0 , and knowing that the discriminant is greater than zero and a perfect square: a. Find k if k is negative. b. Find k if k is positive. The roots are real, rational, and unequal. So, we need two numbers that multiply to 7 that fit that criteria. There are multiple answers, but two options for the roots might be: {-7, -1} or {1, 7} -1 + -7 = -8, so k 8 (the answer to b) 1 + 7 = 8 so k -8 (the answer to a) 6. For the following equation x 2 bx c 0 , if c<0, there are two possibilities for what the roots may be like. Describe these two possibilities and what would have to be true for each situation to occur. Hint: think about the discriminant. If c<0, then b2 – 4ac must be greater than 0 (since a is 1, 4ac will be negative, and if you subtract a negative number, you add the positive number). Then if the discriminant is a perfect square, the roots will be real, rational, and distinct If the discriminant is not a perfect square, the roots will be real, irrational, and distinct 5 7. One root of the equation 3x 2 x k 0 is . Find the other root of the equation. 3 As the equation is written now, we do not know the sum of the roots, since a =/=1. We can divide both sides of the equation by 3. Then, the sum of the roots will be (1/3), since we would see (-1/3) as the middle coefficient. -(5/3) + r = 1/3 R = 6/3 = 2