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Math Help - Help please with a few quadratics

  1. #1
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    Help please with a few quadratics

    I have a few problems I have having difficulty at arriving at a solution that fits correctly.

    The first one solve the equation:

    5x^2 - x - 1 = 0


    The second seems to be very difficult for me:

    The Function
    f (x) = -0.02x^2 + x + 1
    moldels the yearly growth of a young redwood tree, f (x), in inches, with x inches of rainfall per year. How many inches of rainfall per year result in maximum tree growth? What is the yearly growth?

    The Third Problem:
    Solve equation by making an appropriate substitution. When necessary check proposed solutions.

    x^-2 + x^-1 - 56 = 0

    Fourth Problem:

    Express solution to equation in simplified form. Rationalize denominators. Solve equation by the square root property.

    (x - 3)^2 = 20
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  2. #2
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    Quote Originally Posted by kbryant05
    I have a few problems I have having difficulty at arriving at a solution that fits correctly.
    The first one solve the equation:
    5x^2 - x - 1 = 0
    ...
    Hello,

    you only have to use the formula for solving this equation:

    x=\frac{1\pm\sqrt{1-4\cdot 5\cdot (-1)}}{2\cdot 5}

    x=\frac{1\pm\sqrt{21}}{10}

    Quote Originally Posted by kbryant05
    The second seems to be very difficult for me:
    The Function
    f (x) = -0.02x^2 + x + 1
    moldels the yearly growth of a young redwood tree, f (x), in inches, with x inches of rainfall per year. How many inches of rainfall per year result in maximum tree growth? What is the yearly growth?
    Transform this equation into the vertex form of a parabola:
    f (x) = -0.02x^2 + x + 1=-0.02(x^2-50x-50) = -0.02((x^2-50x+625)-625-50) = -0.02(x-25)^2+13.5

    That means: At rainfall of 25'' you'll get the maximum growth of 13.5''

    Quote Originally Posted by kbryant05
    The Third Problem:
    Solve equation by making an appropriate substitution. When necessary check proposed solutions.
    x^-2 + x^-1 - 56 = 0
    with this equation it isn't necessary to use substitution method but because you are forced to do so, here it is:

    Set z=\frac{1}{x}\Longrightarrow x=\frac{1}{z}. Then your equation becomes:

    z^2+z-56=0. Now use the formula for solving this quadratic equation and you'll get Z = -8 or z = 7.

    So x=-\frac{1}{8}\ \vee \ x=\frac{1}{7}

    Quote Originally Posted by kbryant05
    Fourth Problem:
    Express solution to equation in simplified form. Rationalize denominators. Solve equation by the square root property.
    (x - 3)^2 = 20
    I'm not certain what you should do here (I mean I don't understand the words) So I'm going to show you what you can possibly do to get a solution:

    Calculate the square-root on both sides of this equation:

    x-3=-\sqrt{20}\ \vee \ x-3=\sqrt{20}. So you get:

    x=3-\sqrt{20}\ \vee \ x=3+\sqrt{20}

    Greetings

    EB
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  3. #3
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    thanks Earboth

    I'm not quite sure how to solve that one myself the fourth one that is.

    The book came up with the answer for the problem:

    (x - 3 )^2 =20

    answer in book is I can't actually write the way you can at the moment the plus sign has a line under it.
    3 + 2 square root 5

    Here is another one I am having some difficulty with too?

    Determine the constant that should be added to the binomial so that it becomes a perfect square trinomial. Then write and factor the trinomial.

    x^2 + 2 over 5 x

    and this one is being able to solve the equation.
    (x + 2)^2 + 25 = 0

    Thanks for your help it really is appreciated. Some of these are probably really simple to some but really confusing to me.
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  4. #4
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    Quote Originally Posted by kbryant05
    I'm not quite sure how to solve that one myself the fourth one that is.

    The book came up with the answer for the problem:

    (x - 3 )^2 =20

    answer in book is I can't actually write the way you can at the moment the plus sign has a line under it.
    3 + 2 square root 5
    Here is another one I am having some difficulty with too?
    Determine the constant that should be added to the binomial so that it becomes a perfect square trinomial. Then write and factor the trinomial.
    x^2 + 2 over 5 x
    and this one is being able to solve the equation.
    (x + 2)^2 + 25 = 0

    Thanks for your help it really is appreciated. Some of these are probably really simple to some but really confusing to me.
    Hello again,

    The answer in the book and my answer are actually the same:

    x=3\pm\sqrt{20}=3\pm\sqrt{4\cdot 5}=3\pm2\cdot \sqrt{5}

    To your 2nd problem:

    x^2 + {2 \over 5} x+\left( \frac{1}{5} \right)^2=\left(x+\frac{1}{5}\right)^2

    to your 3rd problem:

    (x+2)^2 is greater or equal zero. If you add to a positive number (like the square) another positive number (like the 25) the result cannot be zero. So there exist no real solution with this equation.

    Greetings

    EB
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  5. #5
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    Quote Originally Posted by kbryant05
    The Function
    f (x) = -0.02x^2 + x + 1
    moldels the yearly growth of a young redwood tree, f (x), in inches, with x inches of rainfall per year. How many inches of rainfall per year result in maximum tree growth? What is the yearly growth?
    Earboth did it differently. In school they teach that given,
    y=ax^2+bx+c,a\not = 0 then the max/min is at, -\frac{b}{2a}.
    Thus, given,
    y=-0.02x^2+x+1 we have that,
    a=-0.02,b=1 thus the max/min is at,
    -\frac{1}{2(-.02)}=25, now since the leading coefficient (-0.02) is negative this parabola got a maximum point at x=25.
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  6. #6
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    Question Is this true?

    (x + 2)^2 + 25 = 0

    Is that true that there is no solution for the above equation or is that a mistake? I'm lost within my book lies the answer of -2 + 5i (the plus sign has a line under it)What does that really mean?


    Also one other problem I seriously need some help on.


    A baseball player hits a pop fly into the air. The function
    s(t) = - 16t^2 + 64t + 5

    When does the baseball reach its maximum height? What is that height?
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  7. #7
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    Quote Originally Posted by kbryant05
    (x + 2)^2 + 25 = 0

    Is that true that there is no solution for the above equation or is that a mistake? I'm lost within my book lies the answer of -2 + 5i (the plus sign has a line under it)What does that really mean?

    Let's begin.

     (x+2)^2+25=0 use foil

     x^2+4x+4+25=0 add like terms

     x^2+4x+29=0 now put the numbers into the quadratic formula.

     x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} substitute

     x=\frac{-4\pm\sqrt{4^2-4(1)(29)}}{2(1)} multiply

     x=\frac{-4\pm\sqrt{16-116}}{2} subtract

     x=\frac{-4\pm\sqrt{-100}}{2}

    Now look, there is no square root of -100, so there is no solution. However, you can use imaginary numbers from this point on.

     x=\frac{-4\pm10i}{2} divide

     x=-2\pm5i and here is the answer.

    Quote Originally Posted by kbryant05
    Also one other problem I seriously need some help on.


    A baseball player hits a pop fly into the air. The function
    s(t) = - 16t^2 + 64t + 5

    When does the baseball reach its maximum height? What is that height?
    The maximum height for an upside-down parabola is at its vertex. The equation for a vertex, (as hacker said) is  -\frac{b}{2a} . The vertex is always an x-coordinate, so plug in  -\frac{b}{2a} for x in your equation and you'll find the y-coordinate.
    Last edited by Quick; June 17th 2006 at 10:46 AM.
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  8. #8
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    ...and this one is being able to solve the equation.
    (x + 2)^2 + 25 = 0


    Hi:

    Earboth is absolutely correct in his asserting that (x + 2)^2 + 25 = 0 is without real solutions. There is a nice means by which you can make such determination at a glance. If we consider the function defined by, f(x) = (x + 2)^2 + 25, we immediately observe that f is quadratic with a positive leading coefficient and, therefore, parabolic with upward concavity in the coordinate plane. The expression, presented in "vertex form", gives vertex at (-2, 25). Moreover, the given equation, i.e., (x + 2)^2 + 25 = 0, has solutions which, if real, are located in the x-axis (y=o). But, as we have seen, the graph does not cross the x-axis. [Why?] Hence the equation is without real solutions.


    However, every quadratic equation, indeed every polynomial equation, can be shown to posses a non-empty solution set. Such solution(s) must therefore be imaginary. To that end, (x + 2)^2 + 25 = 0 implies (x + 2)^2 = -25. Hence, sqrt((x + 2)^2) = sqrt(-25). Now,



    Sqrt((x + 2)^2) = |x+2|, and sqrt(-25) = 5i. Thus,
    |x+2| = 5i, ==> x + 2 = +/- 5i, ==> x = -2 +/- 5i.

    The solutions are, therefore, -2+ 5i and -2 + 5i.


    Regards,


    Rich B.
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  9. #9
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    Exclamation Try again

    Okay so I am trying once again to do the baseball equation and I keep trying. I am not mathematically inclined on some of these problems. So forgive me Quick if it seems repetitious to you. Some I can do no problem, others it takes someone else showing me how.

    To all of you Earboth, Hacker and Rb thanks for your kindness and showing me that sometimes it takes just a little help to be able to solve the rest of the challenges in front of me.
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  10. #10
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    kbryant05, Quick's method is not the quickest.

    You may do this,
    (x+2)^2+25=0
    Write as,
    (x+2)^2=-25
    Take square roots,
    x+2=\pm \sqrt{-25}
    Note, that since the radical is negative do it as it it were, \sqrt{25}=5 and attach and "i" thus,
    x+2=\pm 5i
    Subtract two,
    x=-2\pm 5i
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  11. #11
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    Question Earboth or Hacker are you around?

    A baseball player hits a pop fly into the air. The function
    s(t) = - 16t^2 + 64t + 5

    When does the baseball reach its maximum height? What is that height?

    Okay I have tried several times to come up with this answer as in the book which is
    The baseball will reach its maximum height of 69 feet after 2 seconds.

    I am really confused once I turn the equation into the vertex.

    Please help me. I know you both have different ways of doing them and I have tried both but I am completely missing something here?
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  12. #12
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    Quote Originally Posted by kbryant05
    A baseball player hits a pop fly into the air. The function
    s(t) = - 16t^2 + 64t + 5
    When does the baseball reach its maximum height? What is that height?
    ...
    The baseball will reach its maximum height of 69 feet after 2 seconds....
    Hello,

    the graph of this function is a parabola, upside down. So the highest point of this parabola is the vertex. If you transform the equation into vertex form, then you'll get the time and the actual height, when the ball reaches it's peak(?):

    s(t)=-16t^2+64t+5=-16(t^2-4t-\frac{5}{16}). With the first two summands in the bracket you can form a complete square:

    s(t)=-16(t^2-4t-\frac{5}{16})=-16((t^2-4t+4)-4-\frac{5}{16}) = -16((t-2)^2-\frac{69}{16})=-16(t-2)^2+69

    So at t = 2s the ball reaches exactly 69 m.

    Greetings

    EB

    PS: Please do us a favour: If you've got a new problem you need some help with, start a new thread. Otherwise no one will recognize here that you have an other question.
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  13. #13
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    Quote Originally Posted by kbryant05
    A baseball player hits a pop fly into the air. The function
    s(t) = - 16t^2 + 64t + 5

    When does the baseball reach its maximum height? What is that height?
    ...
    Hello,

    it's me again.
    As I've learned from ThePerfectHacker, you maybe are used to do it like this:

    In school they teach that given,
     y=ax^2+bx+c,a\not = 0 then the max/min is at, -\frac{b}{2a}.
    Thus, given,
     y=-16t^2+64t+5 we have that,
    a=-16,b=64 thus the max/min is at: <br />
-\frac{64}{2\cdot (-16)}=2, now since the leading coefficient (-16) is negative this parabola got a maximum point at t=2.

    Plug in this value into the equation and you'll get:

    s(2)=-16\cdot 2^2+64\cdot 2+5=-64+128+5=69

    Greetings

    EB
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  14. #14
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    Wink Thank you Earboth

    Thank you for your help and I actually finally understand better how you did it on your explanation.

    I am still new at posting things, sorry I will in the future post a new thread. Thanks for your advice. Gosh it's amazing how much time one problem can cause a headache for those of us that are confused.
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