Question

SOMEONE PLEASE HELP ME WITH THIS ALGEBRA 2 . its hard and im really stressed

Answer 1

Well, whats the question? :D

Answer 2

Match the complex expression to its correct simplified form.
1. i^22 2. i&7 3. 5 + 8i - 4 + 2i
4. (2 - 3i)(3 + 5i) 5. (2 - i)(2 + i)

Answer 3

i^22 = -1

Answer 4

hmm... well for the first one , we know that
\[i^2= -1 \] so we can split the \[i^{22} \] with the exponent rule
but that can take long... wow @Astrophysics so fast lol

Answer 5

lol . i just dont understand it period

Answer 6

well I know why it was done fast.. the exponent was an even number XD
it's like writing i^2 11 times

Answer 7

Main thing is you understand that i^2 = -1

Answer 8

Which can be proved from \[i^2 = (0+i)^2 \implies (0+i)(0+i) \implies ...-1+0i = -1\]

Answer 9

long way... not recommended lol
\[(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)\]
\[(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)\]
\[(1)(1)(1)(1)(1)(-1) \rightarrow (1)(-1)= -1 \]

Answer 10

lool

Answer 11

Nice haha

Answer 12

i^{100} is gonna hurt by hand XD

Answer 13

so next question
For
3. 5 + 8i - 4 + 2i
we can combine like terms so our answer is in the form of ai+b or a+bi

Answer 14

\[i^7 \implies (i^2)(i^2)(i^2)i\]

Answer 15

so number 2 would be i^7
(-1)(-1)(-1)(-1)(-1)(-1)(-1) = -1

Answer 16

Naw, \[i = \sqrt{-1}\]

Answer 17

oh I read that second problem wrong. I thought it was i and 7 XD

Answer 18

\[i^7 = i^6 i = -1 \times i = -i\]

Answer 19

okay so problem 3 would be i^3 ?

Answer 20

problem 3 has
3. 5 + 8i - 4 + 2i
we can combine like terms so our answer is in the form of ai+b or a+bi

Answer 21

Yeah just combine like terms haha

Answer 22

-1+10i

Answer 23

close but should be +1

Answer 24

we can rearrange it to make it a bit easier
5-4+8i+2i -> 1+10i

Answer 25

well the attempt wasn't that bad.... just 5-4=1

Answer 26

the last 2 problems is just using FOIL and using the fact that\[ i^2 = -1 \] again the final form of these questions is a+bi or ai+b

Answer 27

(2 - i)(2 + i)

4 + 2i - 2i - i^2
4+i^2

Answer 28

did i do that right

Answer 29

\[(2-i)(2+i) \implies (2)(2)+2i+(2)(-i)-i^2 \implies 4+2i-2i-i^2 = 4-i^2 = 4-(-1) = 4+1 = 5\]

Answer 30

\[4-i^2 = 4-(-1) \implies 4+1 = 5\]

Answer 31

(2-i)(2+i)
4+2i-2i-i^2
4-i^2
since i^2 = -1
4-(-1) = 4+1 = 5

Answer 32

Sorry it got cut off there

Answer 33

You're on the right path though, just a sign error :-)

Answer 34

it was a good attempt though .. just when we see i^2, we need to use substitution
Replace i^2 with -1

Answer 35

(2 - 3i)(3 + 5i)
6 + 10i - 9i - 15i
6 - 14i

Answer 36

Hmm I think not

Answer 37

(2-3i)(3+5i)
6+10i-9i-15i^2

Answer 38

remember... i^2 = -1

Answer 39

how do you apply -15 and i^2

Answer 40

\[(2-3i)(3+5i) = 6+10i-9i-15i^2 \implies 6+i-15i^2\]
\[6+i-15(-1) \implies 6+i+15 \implies 21 +i\]

Answer 41

Do you see your error now?

Answer 42

i^2 = -1 replace the i^2 with -1
6+10i-9i-15i^2
since i^2 = -1
6+10i-9i-15(-1)

Answer 43

6+i+15
21+i

Answer 44

\[\huge \checkmark\]

Answer 45

omg thank yall . !!!!!! can you help me with something else ?

Answer 46

1. X^2 + 9 = 0


2. 3x^2 + 11 = -181

Answer 47

are these questions solve for x?

Answer 48

You can set x^2 = -9 this has to do with imaginary numbers as well

Answer 49

Solve the following:

x^2 + 9 = 0

Select one:
a. 2 + 4i
b. + 3i
c. not possible
d. + 9i

Answer 50

Remember as I mentioned before \[i = \sqrt{-1}\]

Answer 51

Solve:
3x^2 + 11 = -181
Select one:
a. not possible
b. + 8i
c. 1 + 13i
d. + 4i

Answer 52

\[x^2 = - 9\] what do you get

Answer 53

ah yes once we square root both sides on x^2=-9
we're gonna have an imaginary case. since negative signs in radicals aren't allowed, the result is imaginary.. denoted by i

Answer 54

3

Answer 55

no

Answer 56

3i ?

Answer 57

That's one solution remember when we take a square root we get a positive and negative value :-)

Answer 58

hmm I guess for this multiple choice, it's only considering the positive version when taking the square root