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Algebra 2 problems

  1. Apr 8, 2004 #1
    Consider the equation y=a(x-r1)(x+r2) for problem 2 & 4.
    Problem 2. State the coordinates of the vertex.
    Problem 4. State the value of the discruminant.

    Problem 14. Solve: sqrt(x-4) + 10 = sqrt(x+4)

    Problem 22.
    Find integers b and c such that the equation x^3+bx^2+cx-10=0 has -2+i as a root.

    Problem 23. If P(x) is a cibic polynomial such that P(-3)=P(-1)=P(2)=0 and P(0)=6, find P(x).
  2. jcsd
  3. Apr 8, 2004 #2
    What progress have you made on the problems so far? Can you show us what you've tried?

  4. Apr 8, 2004 #3
    Problem 18.
    3(2)^2+k(2)-8 ----I inserted the value of x equaling 2.

    For the rectangle problem I believe that the problem is in the ax^2+bx+c=0.

    This is Algebra 2 homework.
  5. Apr 9, 2004 #4
    Regarding Problem 18??

    Sorry it was to be referred to another question I had in a different thread.
  6. Apr 9, 2004 #5


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    (2) Multiply a(x-r1)(x+r2) out and complete the square to find the vertex.

    (4) You got a quadratic when you multiplied in (2) so just use the formula for the discriminant from the quadratic formula.

    (14) Square both sides to get rid of (one of) the square roots. If you still have a square root left, square again!

    (22) Assuming that b and c are supposed to be real numbers, then for the polynomial equation to have -2+i as a root, it must also have -2-i as a root. Now you know a and b in x^3+bx^2+cx-10= (x-a)(x-b)(x-c). Choose c so that -abc= -10.

    (23) Just as the easiest way to solve a polynomial equation is to factor the polynomial, so a cubic with roots a, b, c can be written as (x-a)(x-b)(x-c).
  7. Apr 10, 2004 #6
    For problems 14 and 22.

    Problem 14.
    This is what i have done:
    400(x-4) = 8,464 --I squared both sides.

    Is this right??

    Problem 22.
    This is what I have done:
    x^3+bx^2+cx-10= (x-a)(x-b)(x-c)
    Would i input the values -2+i and -2-i for a and b. TO find the value for c???
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