anonymous
  • anonymous
SOMEONE PLEASE HELP ME WITH THIS ALGEBRA 2 . its hard and im really stressed
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Jhannybean
  • Jhannybean
Well, whats the question? :D
anonymous
  • 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
  • Astrophysics
i^22 = -1

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UsukiDoll
  • 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
  • anonymous
lol . i just dont understand it period
UsukiDoll
  • 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
  • Astrophysics
Main thing is you understand that i^2 = -1
Astrophysics
  • Astrophysics
Which can be proved from \[i^2 = (0+i)^2 \implies (0+i)(0+i) \implies ...-1+0i = -1\]
UsukiDoll
  • 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
  • Astrophysics
lool
Astrophysics
  • Astrophysics
Nice haha
UsukiDoll
  • UsukiDoll
i^{100} is gonna hurt by hand XD
UsukiDoll
  • 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
  • Astrophysics
\[i^7 \implies (i^2)(i^2)(i^2)i\]
anonymous
  • anonymous
so number 2 would be i^7 (-1)(-1)(-1)(-1)(-1)(-1)(-1) = -1
Astrophysics
  • Astrophysics
Naw, \[i = \sqrt{-1}\]
UsukiDoll
  • UsukiDoll
oh I read that second problem wrong. I thought it was i and 7 XD
Astrophysics
  • Astrophysics
\[i^7 = i^6 i = -1 \times i = -i\]
anonymous
  • anonymous
okay so problem 3 would be i^3 ?
UsukiDoll
  • 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
  • Astrophysics
Yeah just combine like terms haha
anonymous
  • anonymous
-1+10i
Astrophysics
  • Astrophysics
close but should be +1
UsukiDoll
  • UsukiDoll
we can rearrange it to make it a bit easier 5-4+8i+2i -> 1+10i
UsukiDoll
  • UsukiDoll
well the attempt wasn't that bad.... just 5-4=1
UsukiDoll
  • 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
  • anonymous
(2 - i)(2 + i) 4 + 2i - 2i - i^2 4+i^2
anonymous
  • anonymous
did i do that right
Astrophysics
  • 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
  • Astrophysics
\[4-i^2 = 4-(-1) \implies 4+1 = 5\]
UsukiDoll
  • UsukiDoll
(2-i)(2+i) 4+2i-2i-i^2 4-i^2 since i^2 = -1 4-(-1) = 4+1 = 5
Astrophysics
  • Astrophysics
Sorry it got cut off there
Astrophysics
  • Astrophysics
You're on the right path though, just a sign error :-)
UsukiDoll
  • UsukiDoll
it was a good attempt though .. just when we see i^2, we need to use substitution Replace i^2 with -1
anonymous
  • anonymous
(2 - 3i)(3 + 5i) 6 + 10i - 9i - 15i 6 - 14i
Astrophysics
  • Astrophysics
Hmm I think not
UsukiDoll
  • UsukiDoll
(2-3i)(3+5i) 6+10i-9i-15i^2
UsukiDoll
  • UsukiDoll
remember... i^2 = -1
anonymous
  • anonymous
how do you apply -15 and i^2
Astrophysics
  • 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
  • Astrophysics
Do you see your error now?
UsukiDoll
  • 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
  • UsukiDoll
6+i+15 21+i
Astrophysics
  • Astrophysics
\[\huge \checkmark\]
anonymous
  • anonymous
omg thank yall . !!!!!! can you help me with something else ?
anonymous
  • anonymous
1. X^2 + 9 = 0 2. 3x^2 + 11 = -181
UsukiDoll
  • UsukiDoll
are these questions solve for x?
Astrophysics
  • Astrophysics
You can set x^2 = -9 this has to do with imaginary numbers as well
anonymous
  • anonymous
Solve the following: x^2 + 9 = 0 Select one: a. 2 + 4i b. + 3i c. not possible d. + 9i
Astrophysics
  • Astrophysics
Remember as I mentioned before \[i = \sqrt{-1}\]
anonymous
  • anonymous
Solve: 3x^2 + 11 = -181 Select one: a. not possible b. + 8i c. 1 + 13i d. + 4i
Astrophysics
  • Astrophysics
\[x^2 = - 9\] what do you get
UsukiDoll
  • 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
  • anonymous
3
anonymous
  • anonymous
no
anonymous
  • anonymous
3i ?
Astrophysics
  • Astrophysics
That's one solution remember when we take a square root we get a positive and negative value :-)
UsukiDoll
  • UsukiDoll
hmm I guess for this multiple choice, it's only considering the positive version when taking the square root |dw:1439181642696:dw|
anonymous
  • anonymous
so positive and negative 3i
Astrophysics
  • Astrophysics
Yup
UsukiDoll
  • UsukiDoll
yeah
Astrophysics
  • Astrophysics
So here is what's going on really
Astrophysics
  • Astrophysics
\[x^2 = -9 \implies x = \pm \sqrt{-9} \implies \pm \sqrt{-1} \sqrt{9} \implies \pm 3i\]
anonymous
  • anonymous
the next one is positive or negative 8i
Astrophysics
  • Astrophysics
We just factor out the \[\sqrt{-1}\]
Astrophysics
  • Astrophysics
Yea thats right
UsukiDoll
  • UsukiDoll
3x^2 +11=-181 3x^2 = -181-11 3x^2=-192 x^2 = -64 X = \sqrt{-64} x = -8i, 8i
anonymous
  • anonymous
(2x^2-8x +10=0) im going to go this one tell me if its correct
Astrophysics
  • Astrophysics
?
anonymous
  • anonymous
|dw:1439181899563:dw|
anonymous
  • anonymous
im not sure what to do next
Astrophysics
  • Astrophysics
Why would you do that, just factor
UsukiDoll
  • UsukiDoll
I wouldn't do that at all
Astrophysics
  • Astrophysics
or complete the square i guess
UsukiDoll
  • UsukiDoll
2x^2-8x+10=0 all of these numbers are even .. .they have one even number in common
Astrophysics
  • Astrophysics
Here lets make it easier on you \[2(x^2-4x+5)=0\]
Astrophysics
  • Astrophysics
Oh sorry Usuki xD
UsukiDoll
  • 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
  • anonymous
then factor x^2 - 4x + 5 ?
UsukiDoll
  • 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
  • Astrophysics
Yup :p
UsukiDoll
  • UsukiDoll
\[\frac{ -b \pm \sqrt{b^2-4ac}}{2a}\]
Astrophysics
  • Astrophysics
Quadratic is the easiest way to approach this
UsukiDoll
  • UsukiDoll
since a = 1, b = -4, and c = 5 but we got our b^2-4ac
Astrophysics
  • Astrophysics
You'll have two complex solutions
UsukiDoll
  • UsukiDoll
\[\frac{ 4 \pm \sqrt{-4}}{2}\] \[\frac{ 4 \pm \sqrt{4}i}{2}\]
UsukiDoll
  • UsukiDoll
take the square root of 4 and then reduce these fractions.
anonymous
  • anonymous
2 plus or minus i
Astrophysics
  • Astrophysics
Nice!
anonymous
  • anonymous
Yay !!! :)
UsukiDoll
  • UsukiDoll
\[\frac{ 4 \pm 2i}{2} \rightarrow 2 \pm i\] correct :)
anonymous
  • anonymous
I have other work to complete if I need help will you all be on ?
UsukiDoll
  • UsukiDoll
depends on the time zone... I'm a bit worn out from last night.
Astrophysics
  • Astrophysics
Just open a new question, and someone will get to it eventually :)
anonymous
  • anonymous
ok thanks for yall help & eastern

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