Evaluate the expression and write the result in the form .
The real number a equals ?
The real number b equals ?
Recall that $\displaystyle i = \sqrt{-1}$, 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."
$\displaystyle \frac{2 + 2i}{1 - 3i} = \frac{2 + 2i}{1 - 3i} \cdot \frac{1 + 3i}{1 + 3i}$
= $\displaystyle \frac{(2 + 2i)(1 + 3i)}{(1 - 3i)(1 + 3i)} = \frac{2 + 6i + 2i + 6i^2}{1 + 3i - 3i - 9i^2}$
= $\displaystyle \frac{2 + 8i - 6}{1 + 9} = \frac{-4 + 8i}{10}$
= $\displaystyle -\frac{4}{10} + i \frac{8}{10} = -\frac{2}{5} + i \frac{4}{5}$
So $\displaystyle a = -\frac{2}{5}$ and $\displaystyle b = \frac{4}{5}$.
-Dan
[quote=badandy328;28053]
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!
$\displaystyle (3 + 5i)(-4 - i)(-1 + 3i) = -(3 + 5i)(4 + i)(-1 + 3i)$
= $\displaystyle -(3 + 5i)(-4 + 12i - i + 3i^2) = -(3 + 5i)(-4 + 11i - 3) = -(3 + 5i)(-7 + 11i)$
= $\displaystyle -(-21 + 33i - 35i + 55i^2) = -(-21 - 2i - 55) = -(-76 - 2i)$
= $\displaystyle 76 + 2i$
$\displaystyle ((-5 - 3i)^2 + 3)i = (25 + 30i + 9i^2 + 3)i = (28 + 30i - 9)i$
= $\displaystyle (19 + 30i)i = 19i + 30i^2 = -30 + 19i$
-Dan
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:
$\displaystyle x^2 - x - 30 = 0$
$\displaystyle (x - 6)(x + 5) = 0$
So the critical points are at x = -5 and x = 6.
Now split the real line into intervals and test each interval.
$\displaystyle (-\infty, -5)$: $\displaystyle x^2 - x - 30 > 0$ Yes!
$\displaystyle (-5, 6)$: $\displaystyle x^2 - x - 30 < 0$ No.
$\displaystyle (6, \infty)$: $\displaystyle x^2 - x - 30 > 0$ Yes!
So the solution set is $\displaystyle (-\infty, -5) \cup (6, \infty)$.
-Dan
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 _________
[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 $\displaystyle a\sqrt{b} + c\sqrt{d}$ is $\displaystyle a\sqrt{b} - c\sqrt{d}$. You only negate one of the terms. Think of it this way. We are applying the identity: $\displaystyle (a + b)(a - b) = a^2 - b^2$. This ensures that the square root signs disappear.
$\displaystyle t + 4 + \frac{6}{t} = 0$
Multiply both sides by t:
$\displaystyle t^2 + 4t + 6 = 0$ <-- We have to remember to check the solutions and make sure $\displaystyle t \neq 0$ because the original expression forbids it. As it happens we won't have to worry here.
Using the quadratic formula:
$\displaystyle t = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}$
$\displaystyle t = \frac{-4 \pm \sqrt{16 - 24}}{2}$
$\displaystyle t = \frac{-4 \pm \sqrt{-8}}{2}$
$\displaystyle t = \frac{-4 \pm \sqrt{-4 \cdot 2}}{2}$
$\displaystyle t = \frac{-4 \pm 2\sqrt{-2}}{2}$
$\displaystyle t = -2 \pm \sqrt{-2}$
$\displaystyle t = -2 \pm i \sqrt{2}$
-Dan
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]