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Quadratic Equantions and Inequalities

  1. Aug 23, 2009 #1
    Hi guys, i'm currently on holiday and my exam starts right after holidays. I bought this book to do some exercise but they only provided answers without solutions. So i wonder if you guys could help me :)

    1. If 3 + [tex]\sqrt{3}[/tex] is a root of the equation 2x2 + 3ax + 3b = 0, where a and b are constants, find the value of ab.

    2. Find the set of values of x such that -16 < x3 - 4x2 + 4x -16 < 0

    3. Show that for all real x, (4x2 + 20x + 19) / (4x + 5) does not lie between 2 and 3.

    I have totally no idea how i should do these. Please help
     
    Last edited: Aug 24, 2009
  2. jcsd
  3. Aug 24, 2009 #2
    Can you show the answers? I'm trying to help
     
  4. Aug 24, 2009 #3
    This is how I would approach problem #1. Factor the original equation into (2X+3)(X+B)=0 you know the first term must be 2X^2 and the last term must be 3B. The middle term must be 3X+2XB=3AX or 3+2B=3A. Plug 3+sqrt(3) into the original equation to get another equation for with A and B. You have two equations and two unknowns. The results for A and B look pretty ugly. I would simplify A, B and X to an approximate number to 4 decimal places and plug the numbers for A,B,and X into the original equation to check if they equal 0.
     
  5. Aug 24, 2009 #4
    Thanks for the help
    the answer as follow:
    1) -16
    2) 0 < x < 4 x cannot be 2
     
  6. Aug 24, 2009 #5
    Are you sure problem #3 is stated correctly? Is it 20^X instead of 20X?
     
  7. Aug 24, 2009 #6
    I'm so sorry! I changed it.
     
  8. Aug 24, 2009 #7
    Problem #3. It looks like the function has a vertical asymptote where the denominator
    4x+5=0. Dividing denominatior into the numerator shows an oblique asymptote at
    y=x+15/4 graphing the function you will see that there are no positive roots.
     
  9. Aug 24, 2009 #8
    In number 2, I think you can use synthetic division to get the values of x.
     
  10. Aug 24, 2009 #9
    Problem #2. X^3-4X^2+4X-16<0 is factorable (X-4)(X^2+4)<0 therefore X<4
    X^3-4X^2+4X>0 is factorable X(X-2)(X-2)>0 therefore X cannot be 0 or 2
    Plugging values into the inequalities around the critical points will give the correct soln. I believe my soln to problem #1 is incorrect. I don't get -16.
     
  11. Aug 24, 2009 #10
    Thanks. I get how to solve #2, i have to split it into two?

    For question #3 i don't get you @_@. The asymptotes shows that it is in x = -5/4
     
  12. Aug 24, 2009 #11
    #1
    [tex]2x^2 + 3ax + 3b = 0[/tex]

    Plug [itex]3+\sqrt{3}[/itex] in the equation.

    Using Vieta's formulas:

    x1+x2=-3a/2

    x1*x2=3b/2

    Solve for x2 and you will get one more equation.

    Solve both equation for a and b, and find a*b.


    #2

    Factor the equation as:

    [tex]-16 < x^2(x-4) + 4(x-4) < 0[/tex]

    [tex]-16< (x-4)(x^2+4) < 0[/tex]

    Now you got separate equations:

    -16<(x-4)(x2+4) & (x-4)(x2+4)<0

    #3

    Find

    [tex]\lim_{x \rightarrow \infty} \frac{4x^2 + 20x + 19}{4x+5}[/tex]

    Regards.
     
  13. Aug 24, 2009 #12
    Thanks for the reply. For Question #3.

    Lim x -> infinite, how am i suppose to do that? My teacher only taught me to divide it with x like if its 5x - 2 then the limit is 5
     
  14. Aug 24, 2009 #13
    Then, divide it by the variable with the biggest power, i.e x2 in this case. What do you get?

    Regards.

    Edit: Here is another approach. #3 Problem says that you need to show that the function doesn't have min. 2 and max.3 for all positive real numbers. Can you do that? You can either use derivatives or you can divide the numerator and denominator and get:

    [tex]\frac{4x^2 + 20x + 19}{4x+5}=\frac{(4x+5)(x+15/4)+1/4}{4x+5}=[/tex]

    Now separate:

    [tex]=x+15/4 + \frac{1}{4(4x+5)}}[/tex]

    What tells you the last equation?

    Since 15/4=3.75, f(x) >3.75, hence you proved that the function does not 2 < f(x) < 3

    Regards.
     
    Last edited: Aug 24, 2009
  15. Aug 24, 2009 #14
    I am confused by the answer to #1. I don't get AB= -16 as stated above. X is given as
    3+sqrt(3). I get A= [(3-2(3+sqrt(3))]/3 and B= -(3+sqrt(3)) AB= 5+3sqrt(3). Plugging X,A and B into the original equation = 0. Am I missing something?
     
  16. Aug 24, 2009 #15

    VietDao29

    User Avatar
    Homework Helper

    This problem cannot be solve since it's lack of information, what kind of number are the constants a, b? Are they real, integer, or natural, or rational, or irrational, or... what? Please copy down the exact problem, or we cannot help you much. :(

    This is pretty easy, you have to solve the system of 2 inequalities:

    [tex]\left\{ \begin{array}{ccc} x ^ 3 - 4x ^ 2 + 4x - 16 & > & -16 \\ x ^ 3 - 4x ^ 2 + 4x - 16 & < & 0 \end{array} \right.[/tex]

    By firstly, do some re-arrangements, to isolate 0 to one side; then factor the other side, and use the table of signs to solve the problem. :)

    I'll give you an example, an easy one, so that hopefully you can remember something:

    Example:

    Solve: x2 + 2x + 3 > 6.

    ----------------------------------

    Subtracting 6 from both sides, yield:

    => x2 + 2x - 3 > 0

    Now, factor the LHS, we have:

    => (x + 3)(x - 1) > 0

    Table of Signs:

    [tex]\begin{tabular}{c|lcccccr} & -\infty & & -3 & & 1 & & +\infty \\ \hline x + 3 & & - & 0 & + & & + \\ \hline x - 1 & & - & & - & 0 & + \\ \hline (x + 3)(x - 1) & & + & 0 & - & 0 & + \end{tabular}[/tex]

    So from the table above, we arrive at the final answer:

    [tex]\left[ \begin{array}{ccc} x & < & -3 \\ x & > & 1 \end{array} \right.[/tex]

    One of the approach is to use proof by contradiction, assume that there exists some [tex]2 < \alpha < 3[/tex], and there is some real number x0, such that:

    [tex]\frac{4x_0 ^ 2 + 20x_0 + 19}{4x_0 + 5} = \alpha[/tex]

    And try to see what contradiction it leads to.

    ---------------------

    Another approach is to find the first, and second derivative of [tex]f(x) = \frac{4x ^ 2 + 20x + 19}{4x + 5}[/tex], and graph it. :)

    ---------------------

    Another way is to use polynomial division, and then use the equality: [tex]a + b \geq 2 \sqrt{ab}[/tex], where a, b > 0.

    [tex]f(x) = \frac{4x ^ 2 + 20x + 19}{4x + 5} = x + \frac{15}{4} + \frac{1}{4(4x + 5)} = \frac{5}{2} + \left( \frac{1}{4} (4x + 5) + \frac{1}{4(4x + 5)} \right)[/tex]

    Remember to split it into 2 cases where 4x + 5 is positive, and 4x + 5 is negative. :)

    Now that I have nothing to say, are you sure f(-1000) > 3.75?!?!?! :|
     
    Last edited: Aug 24, 2009
  17. Aug 24, 2009 #16
    Thanks,
    but after that, what do i do with it? I never heard of the a + b > sqrt of ab before @_@

    Yes i agree with the first one being lack of information, but the problem is, there is 3 similar questions like that. So i thought there must be a way to solve it.

    and Thanks guys, i understand how to do #2 now :)
     
  18. Aug 25, 2009 #17

    VietDao29

    User Avatar
    Homework Helper

    Here's short proof of it. Given a, b >= 0.

    So:
    [tex]a = (\sqrt{a}) ^ 2, b = (\sqrt{b}) ^ 2[/tex]

    Start with:
    [tex]\begin{array}{l} (\sqrt{a} - \sqrt{b}) ^ 2 \geq 0 \\ \Rightarrow (\sqrt{a}) ^ 2 - 2\sqrt{a}\sqrt{b} + (\sqrt{b}) ^ 2 \geq 0 \\ \Rightarrow a - 2\sqrt{ab} + b \geq 0 \\ \Rightarrow a + b \geq 2 \sqrt{ab} \end{array}[/tex]

    -----------------------------------

    Apply this rule to the (...) part, I've grouped that part for you. Just try it, and see how far you can get.

    Remember that the above rule is only applied for a, b >= 0
     
  19. Aug 25, 2009 #18
    Woah, chill down you both @_@. Sorry i still don't seem to get question #3. which is a and which is b ? It doesn't look like a quadratic to me so i don't know which should be a or b.

    or am i suppose to let 4x+5 be something?
     
  20. Aug 25, 2009 #19

    You don't know what you are saying. If vertical asymptote exists then the function is not continuous, so that 2<f(-5/4)<3 is not valid.

    Change signs??

    a=4x
    b=5

    what if x<0, then a<0 so [itex]\sqrt{ab}[/itex] is not in R.

    Please review your knowledge.

    @crays, have you ever learned about derivates?

    Find the roots of f'(x)=0, then substitute them in f''(x), and if f''(x)>0 then the solutions of f'(x) are minimum, and if f''(x)<0 then they are maximum.

    Regards.
     
    Last edited: Aug 25, 2009
  21. Aug 25, 2009 #20

    VietDao29

    User Avatar
    Homework Helper

    What does f(-5/4) has anything to do here? f(-5/4) is clearly not valid, not because there's a vertical asymptote there, but cause the denominator is 0!!!

    What happen when -5/4 < x < 0, and when x < -5/4? What you have done is to proove f(x) > 3.75, when x > 0, and find a vertical asymptote, and.. conclude that the statement of the problem is true for all x? :confused:

    Of course it is not in R, because you haven't changed signs.

    If x < 0 then -x > 0. This is what I call "change signs" :|

    Did I offend you?

    -------------------------

    @OP:
    Well, I'll give you an example, hopefully you can go from here.

    Example:

    Prove the following inequality:

    [tex]\left\{ \begin{array}{ll} x + 1 + \frac{1}{x + 1} \geq 2 & \mbox{when } x > -1, \\ x + 1 + \frac{1}{x + 1} \leq -2 & \mbox{when } x < -1 \right.[/tex]

    • When x > -1, x + 1 is positive, right? And so is 1/(x + 1), using the inequality: [tex]a + b \geq \sqrt{ab}[/tex].

      a = (x + 1), and b = 1/(x + 1), we have:

      [tex]x + 1 + \frac{1}{x + 1} \geq 2 \sqrt{(x + 1) \left( \frac{1}{x + 1} \right)} = 2[/tex].

      Done for x > -1. :)
    • When x < -1, x + 1 is negative, so -(x + 1) is positive? And so is -1/(x + 1), using the inequality: [tex]a + b \geq \sqrt{ab}[/tex].

      a = -(x + 1), and b = -1/(x + 1), we have:

      [tex]\begin{array}{l} -(x + 1) - \frac{1}{x + 1} \geq 2 \sqrt{-(x + 1) \times \left[ -\left( \frac{1}{x + 1} \right) \right]} = 2 \\ \Rightarrow x + 1 + \frac{1}{x + 1} \leq -2 ~ ~ \mbox{multiply both side by -1, and change signs} \end{array}[/tex].

      And so, we've completed our problem. :)

    Your problem can be done in pretty much the same manner. :)

    About problem #1, it clearly is lack of information. You can go back to have a glance at some previous problems of that kind, to see if the books has anything to say about a, and b.
     
    Last edited by a moderator: Aug 25, 2009
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