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For people good in math easy ten points?

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Find the distance between (-2,6) and (1,3)

Find the value of the discriminant of the equation 25x^2 - 30x +9 = 0

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  1. what is the discriminant in the quadratic formula.....its the part under the radical so it's b^2-4(a)(c)

    in this equation a=25 b=30 and c=9

    so just plug the correct values in

    30^2-4(25)(9)

    30^2= 900

    -4(25)= -100

    -100(9)= -900

    so 900-900=0

    what does this tell us.....that the equation has a single real solution  


  2. Distance formula = Sqr root of (x2 - x1)^2 + (y2 - y1)^2

    Given points (-2,6) and (1,3)

    D = Sqr root of (1 - (-2))^2 + (3 - 6)^2 = (3)^2 + (-3)^2 =9 +9 = Sqr root of 18 = Sqr root of 9 x 2 =

    3 radical 2

    The discriminant is part of the quadratic equation.

    The discriminant (D) = b^2 - 4ac

    1)If D > 0, the equation has 2 real solutions.

    2)If D = 0, the equation has 1 real solution.

    3)If D < 0, the equation has 2 conjugate imaginary solutions.

    25x^2 - 30x +9 = 0

    a=25, b= -30, c=9

    b^2 - 4ac = (-30)^2 - 4(25)(9) = 900 - 900 = 0

    Condition #2 is met, so there is just one real solution.

    x = - b / 2a

    x = 30/2(25) = 30/50 = 3/5  

  3. You don't have to be good in Math. You just need to know the formulas.

    for pts (a,b) and (c,d), distance = sqrt[(a-c)^2 + (b-d)^2]

    discriminant of ax^2 + bx + c = 0 is b^2 - 4ac.

  4. using the distance formula,

    d = √(6-3)²+(-2-1)² = √18 = 3√2

    discriminant = b²-4ac = 900-25(9) = 900 - 225 = 675

    that's it! ;)

  5. For the first one, use the distance formula:

    d = sqrt( (x_2 - x_1)^2 + (y_2 - y_1)^2 )

    For the second one, find the discriminant using:

    Delta = b^2 - 4ac

    Your polynomial is: 25x^2 - 30x + 9 = ax^2 + bx + c

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