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Thread: Few question unclear

  1. #1
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    Few question unclear

    a. Simplify 5j(3√-25 -j⁷√-4), giving answer in the form a + bj
    b. 2x + j5 - 6 = jy - j3 + 10, find values x and y
    c. what us the value of x if the function f(x) = 4x - 3x + 1 has the minimum value?
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  2. #2
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    Complex numbers; quadratic expression

    Hello Wakuiz

    Welcome to Math Help Forum!
    Quote Originally Posted by Wakuiz View Post
    a. Simplify 5j(3√-25 -j⁷√-4), giving answer in the form a + bj
    b. 2x + j5 - 6 = jy - j3 + 10, find values x and y
    c. what us the value of x if the function f(x) = 4x - 3x + 1 has the minimum value?
    (a) All you need to do here is to use the fact that $\displaystyle j^2 = -1$ and then use the ordinary rules of algebra.

    So $\displaystyle j^3 = j^2\times j = -j$

    And $\displaystyle j^7 = j^2 \times j^2 \times j^2 \times j = -1\times-1\times-1\times j = -j$

    Also note that $\displaystyle \sqrt{-25} = \sqrt{25}\times \sqrt{-1} = 5j$ and $\displaystyle \sqrt{-4} = 2j$.

    So $\displaystyle 5j^3(3\sqrt{-25} - j^7\sqrt{-4}) = -5j(3\cdot 5j +j\cdot 2j)$

    Can you complete it? I get the final answer $\displaystyle 75+10j$.


    (b) In this question you 'Compare reals and imaginaries'. In other words, the real parts of each side of the equation are equal, and the imaginary parts are equal.

    So, reals: $\displaystyle 2x-6 = 10$

    Imaginaries: $\displaystyle 5 = y-3$

    I'm sure you can solve it now!


    (c) This is all about handling a quadratic expression. You can do it:

    • by differentiating


    • by simply quoting a formula (if you know it) or


    • by completing the square.

    I'll show you the last method, since I'm guessing that might be the way you've been shown.

    $\displaystyle f(x) = 4x^2-3x +1$

    Take out factor $\displaystyle 4$ to make coefficient of $\displaystyle x^2$ equal to $\displaystyle 1$:

    $\displaystyle f(x) = 4(x^2 - \tfrac34x+\tfrac14)$

    Write half the coefficient of $\displaystyle x$ inside a 'bracket squared':

    $\displaystyle f(x) = 4\Big((x-\tfrac38)^2 - ...\Big)$

    Subtract the square of this number, and add in the original constant number:

    $\displaystyle f(x) = 4\Big((x-\tfrac38)^2 - \tfrac{9}{64}+\tfrac14\Big)$

    Simplify:

    $\displaystyle =4\Big((x-\tfrac38)^2 + \tfrac{7}{64}\Big)$

    Since $\displaystyle (x-\tfrac38)^2 \ge 0$ for all real values of $\displaystyle x$, you can then say that the value of $\displaystyle f(x)$ is a minimum when $\displaystyle (x-\tfrac38)^2 = 0$; i.e. when $\displaystyle x = \tfrac38$.

    Grandad
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