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Homework Help: Maths questions

  1. Mar 20, 2004 #1
    I’d really appreciate some help with these problems (they should be fairly easy, I just can’t seem to remember how to do them):

    (1) Find the range of k, element of R, for which x^2 + kx + k = 0 has real roots. If one of the roots of this equation is 2, find the other root.
    (I’ve proved that 0 is greater than or equal to k and 4 is greater than or equal to k, using the b^2-4ac is greater than or equal to 0, but I'm stuck on the last part.)

    (2) Show for any two non zero real numbers, a and b that:
    (b/a^2) + (1/b) is greater than or equal to (-2/a), provided b > 0

    (3) P(z) = z^3 + z, where z is an element of C. Solve the equation P(z) = 0.

    (4) Show that the function f(x) = x^3 + x – 1 has a root between 0 and 1.

    I know it seems like a lot to be asking for help with, but they’re just the questions I'm revising that I can’t do.
    Thanks in advance for any help.
    Last edited: Mar 20, 2004
  2. jcsd
  3. Mar 20, 2004 #2
    (1) If you know that X = 2 solves the equation, substitute 2 into the equation, and find the K for which one of the equation's roots is 2. When you have K, substitute it into the original equation, and you will get a normal quadratic equation. Solve it - one of the roots will be 2, the other one will the answer to the question.
  4. Mar 20, 2004 #3
    (2) Since it's provided that [tex]b > 0[/tex], you can multiply the equation by [tex]b[/tex] without worrying that the sign will change. You can also multiply it by [tex]a^2[/tex], since it's positive always. Move all terms to one side and you will get an interesting equation that can be reduced to something simpler...
  5. Mar 20, 2004 #4
    (3) [tex]P(z) = 0 = z^3 + z = z(z^2 + 1)[/tex]

    One solution is [tex]z = 0[/tex], the other solution is that [tex]z^2 = -1[/tex]. If "z is an element of C" means Z can be a complex number, then you need to take the root of -1 to get the answer for Z. Lucky for you this is a well known number, and you should know that both [tex]i^2[/tex] and [tex](-i)^2[/tex] equal [tex]-1[/tex]. Therefore the solutions for [tex]P(z) = 0[/tex] are:
    [tex]z_1 = 0[/tex]
    [tex]z_2 = i[/tex]
    [tex]z_3 = -i[/tex]

    As for (4), I don't understand how the function can have "a root between 1 and 1". Perhaps you mean between -1 and 1?
  6. Mar 20, 2004 #5
    (4) Ok, I don't know what the correct range is but the solution doesn't really depend on it. I'm assuming you need to prove that the function intersects the X axis somewhere between the points A and B. You start off by finding the derivative of the function, which is:
    [tex]f'(x) = 3x^2 + 1[/tex]
    You can see that the derivative is positive for every x, so the function is going up for every x. Now you need to find the value the function gets for the points A and B. If A < B, and if f(A) < 0 and f(B) > 0, you know for sure that the function intersects the X axis at some point between A and B, since it goes up all the time. Do you see what I mean? :smile:
  7. Mar 20, 2004 #6
    Sorry, (4) was:
    "Show that the function f(x) = x^3 + x – 1 has a root between 0 and 1.", not "1 and 1".

    For (2), i had already tried multipling across by b and a^2 (since a^2 is also known to be positive), but i ended up with "b^2 + 2a + a^2 is greater than or equal to 0". But im not sure how im supposed to draw a conclusion fromt that - maybe if it was "b^2 + 2ab + b^2", but im not sure how to prove it from what I've got...

    Thanks again for your help and fast response.
  8. Mar 20, 2004 #7
    (2) Try multiplying again, you really do get [tex]a^2 + 2ab + b^2 \geq 0[/tex], which is easy to prove. :smile:

    See my last post for a solution for (4). Take A to be 0 and B to be 1.
  9. Mar 20, 2004 #8
    Ok, ill try again, im probably just making some stupid little mistake in my multiplication. Thanks again for the fast help, you've been great.
  10. Mar 20, 2004 #9


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    The easiest way to show that f(x) = x3 + x – 1 has a root between 0 and 1 is to note that f(0)= -1 which is negative and that f(1)= 1+ 1- 1= 1 which is positive.
    Since f is a polynomial it is continuous and so, between x= 0 and x= 1 must take on all values between -1 and 1: in particular 0.

    By the way, most textbooks use the term "root" to mean "solution to an equation" so it would be wrong to talk about a "root of f(x)" but correct to talk about a "root of f(x)= 0". One could as easily talk about "a root of f(x)= 1". Those textbooks use the phrase "a zero of f(x)" to mean "a root of f(x)= 0".
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