a. Simplify 5j³(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) = 4x² - 3x + 1 has the minimum value?
Hello Wakuiz
Welcome to Math Help Forum!(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