I need w/ i^(-15), i^7(1 + i^2), 5x^2 + 2x + 1 = 0, etc.

[hunter23]

It would be great if someone would help me out with these.
I have the answers to them but I just dont know how to solve them myself.

1. i^-15 (an imaginary number with a -15 exponent)

2. i^7(1 + i^2)

3. 5x^2 + 2x + 1 = 0

4. x^4 + 13x^2 + 36 = 0

 

[pka]

\(\displaystyle \begin{array}{l}
i^{ - 1} = - i\quad \& \quad i^n = i^{\bmod (n,4)} \\
\left\{ {\begin{array}{l}
{i^1 = i} \\
{i^2 = - 1} \\
{i^3 = - i} \\
{i^4 = 1} \\
\end{array}} \right. \\
\left( i \right)^{ - 15} = \left( {i^{ - 1} } \right)^{15} \\
\end{array}\)
That is all you need.
mod(n,4) is the remainder when n is divided by 4.
mod(15,4)=3

 

[hunter23]

Thankyou
But um, what do you mean by mod?

 

[pka]

As I wrote above:
mod(n,4) is the remainder when n is divided by 4.
mod(15,4)=3

 

[galactus]

4. \(\displaystyle x^{4} + 13x^{2} + 36 = 0\)

Here's a technique for solving this one. Try it on others.

Ask yourself, "what 2 numbers when multiplied equal 36 and when added equal 13"?.

How about 9 and 4?.

Rewrite by subbing in 9+4 for 13:

\(\displaystyle \L\\x^{4}+9x^{2}+4x^{2}+36\)

\(\displaystyle \L\\(x^{4}+9x^{2})+(4x^{2}+36)\)

Factor:

\(\displaystyle \L\\x^{2}(x^{2}+9)+4(x^{2}+9)\)

\(\displaystyle \L\\(x^{2}+4)(x^{2}+9)=0\)

Now, see what the solutions are?.

It's imperative when you factor that what is inside the parentheses is the same.