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

- anonymous

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

- Stacey Warren - Expert brainly.com

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- schrodinger

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- Jhannybean

Well, whats the question? :D

- anonymous

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)

- Astrophysics

i^22 = -1

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## More answers

- UsukiDoll

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

- anonymous

lol . i just dont understand it period

- UsukiDoll

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

- Astrophysics

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

- Astrophysics

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

- UsukiDoll

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

- Astrophysics

lool

- Astrophysics

Nice haha

- UsukiDoll

i^{100} is gonna hurt by hand XD

- UsukiDoll

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

- Astrophysics

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

- anonymous

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

- Astrophysics

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

- UsukiDoll

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

- Astrophysics

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

- anonymous

okay so problem 3 would be i^3 ?

- UsukiDoll

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

- Astrophysics

Yeah just combine like terms haha

- anonymous

-1+10i

- Astrophysics

close but should be +1

- UsukiDoll

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

- UsukiDoll

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

- UsukiDoll

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

- anonymous

(2 - i)(2 + i)
4 + 2i - 2i - i^2
4+i^2

- anonymous

did i do that right

- Astrophysics

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

- Astrophysics

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

- UsukiDoll

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

- Astrophysics

Sorry it got cut off there

- Astrophysics

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

- UsukiDoll

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

- anonymous

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

- Astrophysics

Hmm I think not

- UsukiDoll

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

- UsukiDoll

remember... i^2 = -1

- anonymous

how do you apply -15 and i^2

- Astrophysics

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

- Astrophysics

Do you see your error now?

- UsukiDoll

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

- UsukiDoll

6+i+15
21+i

- Astrophysics

\[\huge \checkmark\]

- anonymous

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

- anonymous

1. X^2 + 9 = 0
2. 3x^2 + 11 = -181

- UsukiDoll

are these questions solve for x?

- Astrophysics

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

- anonymous

Solve the following:
x^2 + 9 = 0
Select one:
a. 2 + 4i
b. + 3i
c. not possible
d. + 9i

- Astrophysics

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

- anonymous

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

- Astrophysics

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

- UsukiDoll

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

- anonymous

3

- anonymous

no

- anonymous

3i ?

- Astrophysics

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

- UsukiDoll

hmm I guess for this multiple choice, it's only considering the positive version when taking the square root
|dw:1439181642696:dw|

- anonymous

so positive and negative 3i

- Astrophysics

Yup

- UsukiDoll

yeah

- Astrophysics

So here is what's going on really

- Astrophysics

\[x^2 = -9 \implies x = \pm \sqrt{-9} \implies \pm \sqrt{-1} \sqrt{9} \implies \pm 3i\]

- anonymous

the next one is positive or negative 8i

- Astrophysics

We just factor out the \[\sqrt{-1}\]

- Astrophysics

Yea thats right

- UsukiDoll

3x^2 +11=-181
3x^2 = -181-11
3x^2=-192
x^2 = -64
X = \sqrt{-64}
x = -8i, 8i

- anonymous

(2x^2-8x +10=0)
im going to go this one tell me if its correct

- Astrophysics

?

- anonymous

|dw:1439181899563:dw|

- anonymous

im not sure what to do next

- Astrophysics

Why would you do that, just factor

- UsukiDoll

I wouldn't do that at all

- Astrophysics

or complete the square i guess

- UsukiDoll

2x^2-8x+10=0
all of these numbers are even .. .they have one even number in common

- Astrophysics

Here lets make it easier on you \[2(x^2-4x+5)=0\]

- Astrophysics

Oh sorry Usuki xD

- UsukiDoll

2 x 1 = 2
2 x 4 = 8
2 x 5 = 10
all of them have 2 in common . so yank that out and we should have
2(x^2-4x+5) = 0

- anonymous

then factor x^2 - 4x + 5 ?

- UsukiDoll

then factor (x^2-4x+5) since 5 is a prime number the only combination is 1 x 5 = 5
we need a negative number to produce -4
so 1 -5 = -4
(x+1)(x-5) ... only that produces a different sign patern
but I'm wondering if this is written right. by discriminant formula
(-4)^2-4(1)(5) = 16-20 = -4
oh great we need quadratics -_-! .. complex roots.. mhm

- Astrophysics

Yup :p

- UsukiDoll

\[\frac{ -b \pm \sqrt{b^2-4ac}}{2a}\]

- Astrophysics

Quadratic is the easiest way to approach this

- UsukiDoll

since a = 1, b = -4, and c = 5
but we got our b^2-4ac

- Astrophysics

You'll have two complex solutions

- UsukiDoll

\[\frac{ 4 \pm \sqrt{-4}}{2}\]
\[\frac{ 4 \pm \sqrt{4}i}{2}\]

- UsukiDoll

take the square root of 4 and then reduce these fractions.

- anonymous

2 plus or minus i

- Astrophysics

Nice!

- anonymous

Yay !!! :)

- UsukiDoll

\[\frac{ 4 \pm 2i}{2} \rightarrow 2 \pm i\] correct :)

- anonymous

I have other work to complete if I need help will you all be on ?

- UsukiDoll

depends on the time zone... I'm a bit worn out from last night.

- Astrophysics

Just open a new question, and someone will get to it eventually :)

- anonymous

ok thanks for yall help & eastern

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