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Math Help - Quadratic Equation Part 4

  1. #1
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    Quadratic Equation Part 4

    1. Solve (\frac{3-i}{2+i})z=\frac{5+5i}{-1+2i}. Give your answer in form a+bi
    a. Show z on an Argand diagram, find the modulus and argument of z

    2. Find the set of values of x such that \mid \frac{2x+3}{x-1}\mid<1

    3. Given ax^2+bx+c=0. Deduce that the equation has two real roots if b^2-4ac\geq0 and it has two complex roots if b^2-4ac\leq0
    a. Determine all values of k so that x^2-(k-3)x+k^2+2k+5=0 has real roots
    b. If \alpha and \beta are two real roots of this equation, show that <br />
\alpha^2+\beta^2=-(k+5)^2+24. Hence find the maximum value of \alpha^2+\beta^2

    4. What is the difference between \log and \lg?
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  2. #2
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    Quote Originally Posted by cloud5 View Post
    1. Solve (\frac{3-i}{2+i})z=\frac{5+5i}{-1+2i}. Give your answer in form a+bi
    a. Show z on an Argand diagram, find the modulus and argument of z
    (1) z=\frac{5+5i}{-1+2i}\cdot \frac{2+i}{3-i}

    =\frac{10+5i+10i+5i^2}{-3+i+6i-2i^2}

    =\frac{5+15i}{-1+7i}

    Then multiply by its conjugate .

    =\frac{5+15i}{-1+7i}\cdot \frac{-1-7i}{-1-7i}

    =\frac{100-50i}{50}

    Therefore , z=2-i

    |z|=\sqrt{2^2+(-1)^2}

    \tan \theta=\frac{1}{2}

    \theta=26.57

     <br />
\arg z=-26.57 <br />
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  3. #3
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    Quote Originally Posted by cloud5 View Post
    1. Solve (\frac{3-i}{2+i})z=\frac{5+5i}{-1+2i}. Give your answer in form a+bi
    2. Find the set of values of x such that \mid \frac{2x+3}{x-1}\mid<1

    [/tex]

    4. What is the difference between \log and \lg?
    (2) \frac{|2x+3|}{|x-1|}<1

    |2x+3|<|x-1| , square both sides

    4x^2+12x+9<x^2-2x+1

    3x^2+14x+8<0

    (3x+2)(x+4)<0

    Carry on ..

    (4) lg is another way of writing log_{10}

    In other words , lg refers to only log to the base 10

    log can be log to the base 2 or 3 or 4 ..
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  4. #4
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    Quote Originally Posted by cloud5 View Post



    3. Given ax^2+bx+c=0. Deduce that the equation has two real roots if b^2-4ac\geq0 and it has two complex roots if b^2-4ac\leq0
    a. Determine all values of k so that x^2-(k-3)x+k^2+2k+5=0 has real roots
    b. If \alpha and \beta are two real roots of this equation, show that <br />
\alpha^2+\beta^2=-(k+5)^2+24. Hence find the maximum value of \alpha^2+\beta^2
    ax^2+bx+c=0

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

    When b^2-4ac\geq 0

    What can you deduce and when its \leq what can you deduce ?

    (a) Solve [-(k-3)]^2-4[k^2+2k+5]\geq 0

    (b) Let the 2 roots be p and q .

    p+q=k-3 , pq=k^2+2k+5

    (p+q)^2-2pq=(k-3)^2-2[k^2+2k+5]

    =-k^2-10k-1

    =-(k^2+10k+1)

    =-[(k+5)^2-24]

    =-(k+5)^2+24

    For the max , observe the range of the real roots , sketch the graph , and take the one nearest to 24 .
    Last edited by mathaddict; November 26th 2009 at 02:22 AM.
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