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Thread: some simple questions.from my exam(roots ,partial fractions ..)

  1. #1
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    some simple questions.from my exam(roots ,partial fractions ..)

    hi guys i had an exam today and just wanted you guys to give me the correct answer for it so i can see how i went.
    i'm sure its really easy for you guys

    1) if $\displaystyle \alpha $and$\displaystyle \beta$ are the roots of the equation x^2 - 5x + 4 = 0 find
    i) $\displaystyle \alpha^2 + \beta^2$ ii) $\displaystyle \alpha^2 \beta^2$ iii) the equation with roots $\displaystyle \alpha^2$ and $\displaystyle \beta^2$


    2) the curve$\displaystyle f(x) = ax^2 + bx + c$ has factors$\displaystyle (x + 1)$ and$\displaystyle (x + 3) $and cuts the y-axis at $\displaystyle y = 8$ . Find the values of a,b,c

    3) Express in partial fractions $\displaystyle \frac{2}{(x-2)(2x^2+1)}$ in partial fractions.

    4) Simplify $\displaystyle \frac{1}{\sqrt{x+1}}+\sqrt{x}$, rationalising the denominator.

    thanks for the help!
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  2. #2
    Super Member dhiab's Avatar
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    Hello : Question 1


    the equation is :
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  3. #3
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    Hello, llkkjj24!


    1) if $\displaystyle \alpha$ and $\displaystyle \beta$ are the roots of the equation: $\displaystyle x^2 - 5x + 4 \:=\: 0$, find:

    $\displaystyle (i)\;\;\alpha^2 + \beta^2$
    We know that: .$\displaystyle \begin{array}{cc}\alpha + \beta \:=\:5 & [1] \\ \alpha\!\cdot\!\beta \:=\:4 & [2] \end{array}$


    $\displaystyle \text{Square [1]: }\;(\alpha + \beta)^2 \:=\:5^2 \quad\Rightarrow\quad \alpha^2 + 2(\alpha\beta) + \beta^2 \:=\:25$
    . . . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle \uparrow$
    . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle ^{\text{This is 4}}$

    And we have: .$\displaystyle \alpha^2 + 8 + \beta^2 \:=\:25 \quad\Rightarrow\quad \alpha^2 + \beta^2 \:=\:17$




    $\displaystyle (ii)\;\;\alpha^2 \beta^2$
    Since $\displaystyle \alpha\beta \:=\:4$, then: .$\displaystyle \alpha^2\beta^2 \:=\:16$



    $\displaystyle (iii)$ the equation with roots $\displaystyle \alpha^2$ and $\displaystyle \beta^2$
    The equation is: .$\displaystyle (x - \alpha^2)(x - \beta^2) \:=\:0 \quad\Rightarrow\quad x^2 - \alpha^2x - \beta^2x + \alpha^2\beta^2\:=\:0 $

    . . $\displaystyle x^2 - \underbrace{(\alpha^2 + \beta^2)}x + \underbrace{\alpha^2\beta^2} \:=\:0$
    . . . . . . . .$\displaystyle \uparrow$ . . . . . .$\displaystyle \uparrow$
    . . . . . .$\displaystyle ^{\text{This is 17}}$ . . $\displaystyle ^{\text{This is 16}}$


    Therefore: .$\displaystyle x^2 - 17x + 16 \;=\;0$

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  4. #4
    Super Member dhiab's Avatar
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    Quote Originally Posted by llkkjj24 View Post
    hi guys i had an exam today and just wanted you guys to give me the correct answer for it so i can see how i went.
    i'm sure its really easy for you guys

    1) if $\displaystyle \alpha $and$\displaystyle \beta$ are the roots of the equation x^2 - 5x + 4 = 0 find
    i) $\displaystyle \alpha^2 + \beta^2$ ii) $\displaystyle \alpha^2 \beta^2$ iii) the equation with roots $\displaystyle \alpha^2$ and $\displaystyle \beta^2$


    2) the curve$\displaystyle f(x) = ax^2 + bx + c$ has factors$\displaystyle (x + 1)$ and$\displaystyle (x + 3) $and cuts the y-axis at $\displaystyle y = 8$ . Find the values of a,b,c

    3) Express in partial fractions $\displaystyle \frac{2}{(x-2)(2x^2+1)}$ in partial fractions.

    4) Simplify $\displaystyle \frac{1}{\sqrt{x+1}}+\sqrt{x}$, rationalising the denominator.

    thanks for the help!
    HELLO : QUESTION 2
    Ihave the system
    Continu......
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