1. ## complex numbers

Evaluate the expression and write the result in the form .

The real number a equals ?
The real number b equals ?

2. Need help with these to:
The expression

equals where
the coefficient C is _______ , the exponent e is______

Write the following numbers in a+b form:

Solve the following inequality. Write the answer in interval notation.

Evaluate the expression and write the result in the form .

The real number a equals ?
The real number b equals ?
Recall that $i = \sqrt{-1}$, so basically this is a problem in rationalizing the denominator. As in those types of problems, we want to multiply the top and bottom by the "conjugate" of the denominator, in this case called the "complex conjugate."

$\frac{2 + 2i}{1 - 3i} = \frac{2 + 2i}{1 - 3i} \cdot \frac{1 + 3i}{1 + 3i}$

= $\frac{(2 + 2i)(1 + 3i)}{(1 - 3i)(1 + 3i)} = \frac{2 + 6i + 2i + 6i^2}{1 + 3i - 3i - 9i^2}$

= $\frac{2 + 8i - 6}{1 + 9} = \frac{-4 + 8i}{10}$

= $-\frac{4}{10} + i \frac{8}{10} = -\frac{2}{5} + i \frac{4}{5}$

So $a = -\frac{2}{5}$ and $b = \frac{4}{5}$.

-Dan

equals where
the coefficient C is _______ , the exponent e is______
$\frac{x^7(2x)^7}{x^2} = \frac{x^22^7x^7}{x^2}$

= $2^7 \frac{x^{7+7}}{x^2} = 128\frac{x^{14}}{x^2}$

= $128x^{14-2} = 128x^{12}$

So C = 128 and e = 12.

-Dan

Write the following numbers in a+b form:

[quote]
I'm going to FOIL out each pair of factors, but I'm going to factor a "-" from the second term. This is simply for book-keeping: I hate keeping track of negative signs!

$(3 + 5i)(-4 - i)(-1 + 3i) = -(3 + 5i)(4 + i)(-1 + 3i)$

= $-(3 + 5i)(-4 + 12i - i + 3i^2) = -(3 + 5i)(-4 + 11i - 3) = -(3 + 5i)(-7 + 11i)$

= $-(-21 + 33i - 35i + 55i^2) = -(-21 - 2i - 55) = -(-76 - 2i)$

= $76 + 2i$

$((-5 - 3i)^2 + 3)i = (25 + 30i + 9i^2 + 3)i = (28 + 30i - 9)i$

= $(19 + 30i)i = 19i + 30i^2 = -30 + 19i$

-Dan

6. Thank you sooo much. Is there like a site where you plug this stuff in or are you just actually solving them?

Solve the following inequality. Write the answer in interval notation.
First look for the critical points. This is where the expression is 0 or where a 0 appears in the denominator. Since we have no denominator, all we need to know is where the expression is 0. So:
$x^2 - x - 30 = 0$

$(x - 6)(x + 5) = 0$

So the critical points are at x = -5 and x = 6.

Now split the real line into intervals and test each interval.
$(-\infty, -5)$: $x^2 - x - 30 > 0$ Yes!
$(-5, 6)$: $x^2 - x - 30 < 0$ No.
$(6, \infty)$: $x^2 - x - 30 > 0$ Yes!

So the solution set is $(-\infty, -5) \cup (6, \infty)$.

-Dan

Thank you sooo much. Is there like a site where you plug this stuff in or are you just actually solving them?
No. I'm solving them. Though I admit to checking the answer on my calculator to help speed the process of finding my mistakes before I post.

-Dan

9. 2 last things. Is this right?
If you rationalize the denominator of

then you will get

where r, s, and n are all positive integers (with no common factors).
r= -7
s= 4
n=-118

And the last question I have no clue how to do.
Find all solutions of the equation and express them in the form :
First input the solution with b<0 here:
the real number a equals _____ and the real number b equals ______
Then input the solution with b>0 here:
the real number a equals _______ and the real number b equals _________

10. [quote=badandy328;28061]2 last things. Is this right?
If you rationalize the denominator of

then you will get

where r, s, and n are all positive integers (with no common factors).
r= -7
s= 4
n=-118
[\quote]

Almost. The conjugate of $a\sqrt{b} + c\sqrt{d}$ is $a\sqrt{b} - c\sqrt{d}$. You only negate one of the terms. Think of it this way. We are applying the identity: $(a + b)(a - b) = a^2 - b^2$. This ensures that the square root signs disappear.

And the last question I have no clue how to do.
Find all solutions of the equation and express them in the form :
First input the solution with b<0 here:
the real number a equals _____ and the real number b equals ______
Then input the solution with b>0 here:
the real number a equals _______ and the real number b equals _________
$t + 4 + \frac{6}{t} = 0$

Multiply both sides by t:
$t^2 + 4t + 6 = 0$ <-- We have to remember to check the solutions and make sure $t \neq 0$ because the original expression forbids it. As it happens we won't have to worry here.

$t = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}$

$t = \frac{-4 \pm \sqrt{16 - 24}}{2}$

$t = \frac{-4 \pm \sqrt{-8}}{2}$

$t = \frac{-4 \pm \sqrt{-4 \cdot 2}}{2}$

$t = \frac{-4 \pm 2\sqrt{-2}}{2}$

$t = -2 \pm \sqrt{-2}$

$t = -2 \pm i \sqrt{2}$

-Dan

11. Ok, I see how to came to figure that out. Now I'm having trouble figuring out how to answer that question.

Ok, I see how to came to figure that out. Now I'm having trouble figuring out how to answer that question.
Which question?

-Dan

13. This one:

And the last question I have no clue how to do.
Find all solutions of the equation and express them in the form :
First input the solution with b<0 here:
the real number a equals _____ and the real number b equals ______
Then input the solution with b>0 here:
the real number a equals _______ and the real number b equals _________[/QUOTE]

This one:

And the last question I have no clue how to do.
Find all solutions of the equation and express them in the form :
First input the solution with b<0 here:
the real number a equals _____ and the real number b equals ______
Then input the solution with b>0 here:
the real number a equals _______ and the real number b equals _________
Ummmm...I did post the solution...

The real number a is -2 for both questions and b is $\pm \sqrt{2}$.

-Dan