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Math Help - Please help me check these problems (discriminants, complex solutions, inequalities)

  1. #1
    Member
    Joined
    Aug 2009
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    113

    Please help me check these problems (discriminants, complex solutions, inequalities)

    Thanks will be given to those who help. I am in dire need of it too, for some problems. Here is my work:



    1.Find the discriminant


    -10x^2 - 6x + 5 = 0


    -(-6)^2 -4(-10)(5)
    (36 + 200)
    236


    236 is the discriminant.


    2. Find the discriminant


    2x^2 + 6x + 1 = 0


    (6)^2- 4(2)(1)
    (36 - 8)
    28


    28 is the discriminant


    3. Could someone continue this and show me the complex solution, I
    cannot calculate it and have been struggling with the complex number part for a long time. Here is what I have so far:


    2x^2 - 5x + 4 = 0

    x = -(-5)+_ SQRT((-5)^2 -4(2)(4))/2(2)
    = 5 +_ SQRT(25 - 32)/4 = 5 +_SQRT(-7)/4


    4. Could someone please show me how to solve these two (#4/#5)?


    4x^2 - 5x - 6 <= 0



    5. 2x^2 + 9x + 4 < 0


    6. x^2 + 3x - 10 = 0 This one was solved using the quadratic formula, but I need to find the pair of numbers for: The product of two numbers is 10. One number is 3 more than the other number. What is the quadratic equation that models these numbers? How would I do this?


    x = -(3) +_ SQRT((3)^2) - 4(1)(-10)/2(1)
    = -3 +_ SQRT(9 + 40)/2 = -3 +_ SQRT(49)/ 2
    = -3 +_7/2 = -3 - 7/2, -3 + 7/2
    = -10/2, 4/2 = -5, 2
    x = -5, x = 2


    Thanks!
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  2. #2
    Senior Member
    Joined
    Jul 2009
    Posts
    397
    Hi bobbyboy1111

    1. correct
    2. correct

    3. \frac{5\pm\sqrt{-7}}{4}=\frac{5\pm i\sqrt{7}}{4}

    Note that : i=\sqrt{-1}


    4 and 5. You can start by factorizing them and find the critical values


    6. Another way to solve x^2 + 3x - 10 = 0 :

    x^2 + 3x - 10 = 0

    (x+5) (x-2)=0

     x=-5 , x = 2

    Then, you need to find the pair of numbers for: The product of two numbers is 10. One number is 3 more than the other number.

    Let : the numbers are a and b

    *The product of two numbers is 10 -----------------------> ab = 10 ..........(1)
    *One number is 3 more than the other number --------> a = b + 3 ........(2)

    Substitute the two equations. ^^

    EDIT : I think it's better for you to start learning latex : http://www.mathhelpforum.com/math-he...-tutorial.html
    Last edited by songoku; August 11th 2009 at 03:18 AM.
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