Algebra 2

1. The roots of the quadratic equation z^2 + az + b = 0 are 2 - 3i and 2 + 3i. What is a+b?

2. Find all pairs of real numbers (x,y) such that x + y = 6 and x^2 + y^2 = 28. If you find more than one pair, then list your pairs in order by increasing x value, separated by commas. For example, to enter the solutions (2,4) and (-3,9), you would enter "(-3,9),(2,4)" (without the quotation marks).

3. For what real values of a does the quadratic 4x^2 + ax + 25 have nonreal roots? Give your answer as an interval.

4.Find one pair of real numbers, (x,y), such that x + y = 6 and x^3 + y^3 = 144.

5. For what real values of c is x^2 + 16x + c the square of a binomial? If you find more than one, then list your values separated by commas

6. For what real number k does the product (25 + ki)(3+2i) equal a real number?

7. What constant must we place in the blank below in order to be able to factor the resulting expression into the product of two linear factors:
2mn - 16m - 7n +__

8. What number can we place in the blank to make the quadratic below the square of a binomial:
x^2 + (5/3)x +__

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  1. What are you stuck on?
    I would rewrite the first one as
    1 z^2 + b z + c = 0

    z = [-b +/- sqrt(b^2-4c)]/2

    [z+{b -sqrt(b^2-4c)}][z+{b +sqrt(b^2-4c)}]=0
    so
    -b/2=2 so b = -4
    .5 sqrt(16-4c) = 3i
    sqrt (16-4c) = 6i
    16-4c = -36
    4c = 52
    c = 13
    z^2 -4z + 13

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  2. Well I got 9 for the first one, couldn't figure out the 2nd one, {6,0} for the 3rd one, (1,3) for the fourth one, -9,9 for the fifth one, 4 for the 6th one, 3 for the 7th one, and 16 for the last one.
    I don't think I do too well when it gets to 2 variables.

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  3. I like this assignment. Of course, as Damon hinted at, we are not going to do this for you.

    The questions require a basic understanding of these topics.
    Where are your difficulties ?

    Some hints:
    #1 , see Damon's solution
    or, alternate way:
    sum of roots = 2 - 3i + 2 + 3i = 4
    product of roots = (2 - 3i)(2 + 3i) = 4 - 9i^2 = 13
    in z^2 + az + b = 0
    sum of roots = -a/1 , so a = -4
    product of roots = c/1, so c = 13
    a+b = -4+13 = 9

    #2 , y = 6-x
    plug into the circle equation, and solve as a quadratic , hint:use the formula

    #3 properties of the discriminant

    #4, neat question, here is a short cut
    x+y = 6
    cube it
    (x+y)^3 = 216
    x^3 + 3x^2 y + 3x y^2 + y^3 = 216
    x^3+y^3 + 3xy(x+y) = 216
    144 + 3xy(6) = 216
    18xy = 72
    xy = 4

    x(6-x) = 4
    x^2 - 6x = -4
    complete the square, instead of the formula
    x^2 - 6x + 9 = -4+9
    (x-3)^2 = 5
    x-3 = +-SQRT(5)
    x = 3 +-SQRT(5)
    plug into y = 6-x to find the matching y values

    #5 properties of perfect squares, I just used them in #4 when I completed the squrare

    #6. expand it to get a quadratic in terms of k.
    Set the discriminant equal to zero and solve for k

    #7 2mn - 16m - 7n +__
    = 2m(n - 8) - 7(n + ????)

    #8 same type as #5

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