1. ## Few question unclear

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?

2. ## Complex numbers; quadratic expression

Hello Wakuiz

Welcome to Math Help Forum!
Originally Posted by Wakuiz
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?
(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 $\displaystyle x = \tfrac38$.