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Math Help - Pre-Calculus - Quad/Poly Equations and Circles

  1. #1
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    Pre-Calculus - Quad/Poly Equations and Circles

    1. If f(x) = 3x^2 + 5x, then find the value of x such that f(x) = 2
    a) -2 b) -2/3 c) 0 d) 1/3 e)-2

    My work:
    2 = 3x^2+5x
    0 = 3x^2+5x-2
    (Input in graphing calculator, Then find the zeros)
    I got -2 and 1/3

    But the answer is "d". I'm confused.


    2. Solve: 3t^2 + 5t - 12 = 0. How do you solve this algebraically (without using a graphing calculator)? I'm clueless, so I can't show work.
    Answer: t = -3 or 4/3

    Can somebody help me out with these questions?

    AlphaRock

    I have a third one:


    3. If (m, -2) is on a circle with center (-2,2) and a radius 9. Find the value of m.

    My work:
    (x+2)^2 + (-2-2)^2 = 9^2
    (x+2)^2 + (-4)^2 = 81.
    x^2 + 4x + 4 + 16 = 81.
    x^2 + 4x + 20 -20 = 81 - 20
    x^2 + 4x = 61.
    x^2 + 4x - 64 = 0

    Use graphing calculator...
    x = 6.25
    x = -10.25

    REAL ANSWER: -2 +/- sqrt(65)

    Can somebody show me what I can do to make this right?
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  2. #2
    o_O
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    Quote Originally Posted by AlphaRock View Post
    1. If f(x) = 3x^2 + 5x, then find the value of x such that f(x) = 2
    a) -2 b) -2/3 c) 0 d) 1/3 e)-2

    My work:
    2 = 3x^2+5x
    0 = 3x^2+5x-2
    (Input in graphing calculator, Then find the zeros)
    I got -2 and 1/3

    But the answer is "d". I'm confused.

    You're right. Both (d) and (e) work. Just plug them into f(x) and you should get 2.

    2. Solve: 3t^2 + 5t - 12 = 0. How do you solve this algebraically (without using a graphing calculator)? I'm clueless, so I can't show work.
    Answer: t = -3 or 4/3

    Can somebody help me out with these questions?

    AlphaRock
    Do you know how to factor trinomials?

    I have a third one:


    3. If (m, -2) is on a circle with center (-2,2) and a radius 9. Find the value of m.
    In general: (x - a)^2 + (y - b)^2 = r^2 is an equation of a circle with (a,b) as the centre and r as the radius.

    So, we can deduce that your circle is given by:
    \begin{array}{rcl}(x - (-2))^2 + (y - 2)^2  & = & 9^2 \\ (x+2)^2 + (y-2)^2 & = & 81 \end{array}

    Since (m, -2) is on the circle, simply plug them in:
    \begin{array}{rcl} (m + 2)^2 + (-2 - 2)^2 & = & 81 \\ (m+2)^2 +(-4)^2 & = & 81 \\ & \vdots & \end{array}

    And solve.
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  3. #3
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    Quote Originally Posted by o_O View Post

    Do you know how to factor trinomials?

    Thanks, o_O

    I could also use the quadratic formula.

    There is a third way, too. It's to use y = a(x-p)^2 + q.

    What does "p" stand for in that equation...? And how do I get "p"?
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  4. #4
    o_O
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    p is the horizontal shift from the origin. In order to get it in that form, you need to complete the square. However, this will not help you since (p, q) represents the vertex, not the zeroes.

    Factoring would be the quickest I'd say. But the quadratic formula works as well, especially if you have a calculator with this built-in function.
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  5. #5
    Super Member 11rdc11's Avatar
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    In general, can be factored by finding two numbers that multiply to get and add up to , call them . Then you split your quadratic: and you should be able to factor from there (assuming it was factorable in the first place)

    That statement came from a post by o_O

    3t^2 + 5t - 12

    3t^2 +9t -4t -12

    3t(t + 3) -4(t + 3)

    (3t - 4)(t + 3)

    Thanks o_O
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