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Thread: Quadratic Equation Part 4

  1. #1
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    Quadratic Equation Part 4

    1. Solve $\displaystyle (\frac{3-i}{2+i})z=\frac{5+5i}{-1+2i}$. 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 $\displaystyle \mid \frac{2x+3}{x-1}\mid<1$

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

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

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

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

    Then multiply by its conjugate .

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

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

    Therefore , $\displaystyle z=2-i $

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

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

    $\displaystyle \theta=26.57 $

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

    [/tex]

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

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

    $\displaystyle 4x^2+12x+9<x^2-2x+1$

    $\displaystyle 3x^2+14x+8<0$

    $\displaystyle (3x+2)(x+4)<0$

    Carry on ..

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

    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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    Quote Originally Posted by cloud5 View Post



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

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

    When $\displaystyle b^2-4ac\geq 0 $

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

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

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

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

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

    $\displaystyle =-k^2-10k-1$

    $\displaystyle =-(k^2+10k+1)$

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

    $\displaystyle =-(k+5)^2+24$

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