# Questions on Algebra: Complex Numbers answered by real tutors!

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 Algebra: Complex Numbers Solvers Lessons Answers archive Quiz In Depth

Question 751680: Use the quadratic formula to solve the quadratic equation x^2 − 8x − 4 = 0.

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 Solved by pluggable solver: SOLVE quadratic equation with variable Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=80 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 8.47213595499958, -0.47213595499958. Here's your graph:

Question 751571: xsquared + 10x + 2x

Question 751572: xsquared - 8x + 12

Question 751397: How do you solve
3-4i-(-5+2i)?

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Distribute and Combine like terms

3-4i-(-5+2i)

3-4i+5-2i

(3+5)+(-4i-2i)

8 - 6i

Question 750417: (-8-6i)-(5+3i)

Question 749966: Bob and Alice can finish a job together in 3 hours. If Bob can do the job by himself in 5 hours, what percent of the job does Alice do?
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This describes three rates and two of them are related.

Bob's rate----------------- jobs per hour
Alice's rate-----------------r, we do not know but wish to find
Bob and Alice Together, rate-------- jobs per hour

The rates are shown as "jobs per hour" because in this way, two people working together makes their rates additive.
This is what that means:
Bob's rate + Alice's rate = The combined rate of Bob and Alice

Solve for r. Reminder, r will be in units of jobs per hour.

Question 728671: Two complex numbers: a + bi and c + di are equal if a = c and b = d. Use that fact to solve for x and y : ( 3x -yi ) i = 6 ( 1 + yi)
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( 3x -yi ) i = 6 ( 1 + yi)
3xi+y = 6 +6yi
y+3xi=6+6yi
real:
y=6
Imaginary:
3x=6y=6(6)
x=12

:)

Question 733134: Find the complex zeros of each polynomial function. write f in factored form.
f(x)=x^4+ 2x^3 + 22x^2 +50x- 75

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x^4+ 2x^3 + 22x^2 +50x- 75=
(2x^3 +50x)+(x^4 + 22x^2 - 75)=
2x(x^2 +25)+(x^4 + 25x^2 -3x^2- 75)=
2x(x^2 +25)+(x^4 + 25x^2) -(3x^2+ 75)=
2x(x^2 +25)+x^2(x^2 + 25) -3(x^2+ 25)=
(x^2 +25)(x^2+2x-3)=
(x^2+25)(x+3)(x-1)

zeros: {-5i, 5i, -3, 1}

:)

Question 740801: What are 4'th roots of -8i
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-8i = 8 at 270 deg
n=4
k=0,1,2,3
k=0:
r= 4-th-root(8)at theta=(270)/4=135/2
k=1:
r=4-th-root(8) at theta=(270+360)/4=315/2
k=2:
r=4-th-root(8) at theta=(270+720)/4=495/2
k=3:
r=4-th-root(8) at theta=(270+1080)/4=675/2

:)

Question 738211: express in polar form:
a) 2i
b) -3

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a) 2i = 2 at 90 degrees
b) -3 = 3 at 180 degrees

:)

Question 743739: Find the equation of the tangent line for f(x)=3x^2-x at the given point (-2,14)
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f' = 6x-1
f'(-2)=6(-2)-1=-13

so m=-13
y-b=m(x-a)
y-14=-13(x--2)
y-14=-13(x+2)

:)

Question 733131: Find the remaining zeros of each function. h(x)= 3xh(x)= 3x^5 +2x^4+ 15x^3 + 10x^2- 582x- 352, zero; -4i.
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I checked with software and -4i is not a zero for this polynomial, please repost :)

Question 748147: Solve using the quadratic formula: 5x^2 − 8x − 3 = 0

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Solve using the quadratic formula: 5x^2 − 8x − 3 = 0
(-b+or-sqrt(b^2-4ac)/2a)
(8+or-sqrt(64-4*5*(-3))/2*5
(8+or-sqrt(64+60)/10
(8+or-sqrt(124))/10
x=(8+12)/10=2
x=(8-12)/10=-4/10=-2/10=-.2

Question 747939: Multiply
(9+9i)(7+i)Can someone please help me with how to do the FOIL on this? I tried and got it messed up. Thanks for the Help :)

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63+9i+63i+9i^2=
63+72i=9i^2

Question 747934: Subtract and simplify
(4-i)-(7+4i)
I know that I need to use FOIL but I get the answer all wrong. Can someone help me understand?
Thanks,
Cassie

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Subtract and simplify
(4-i)-(7+4i)
I know that I need to use FOIL but I get the answer all wrong.
----
No, FOIL is used to multiply two binomials.
Yours is a subtraction problem.
-----
= 4-i-7-4i
----
= -3-5i
=============
Cheers,
Stan H.

Question 747794: 5^(3x+8)=7
(five to the power of 3x+8) please explain aswell

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5^(3x+8)=7
--------------
(3x+8)*log(5) = log(7)

Question 741034: if 3^x=10 is solved for x, which of the following is the solution rounded to the hundreths place?
a. 2.10
b. 2.09
c. 2.19
d. 2.08

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X= 1/.477 = 2.0959 rounds to 2.1

Question 747273: Will you please show me how to write in a+bi form?
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There appears to be no real portion, only a coefficient of i

Question 746845: Degree 6;zeros:-4,6,4-5i,-6+i
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if you mean find a polynomial of degree 6 with those zeros then here goes...

the general form would be:

which after some work we get:

and the principal polynomial is:

Question 740630: Simplify:

The answer should be in the form of

Question 739005: Hello!!
(2x-1)+5i=9+(y-1)i
I'm a bit confused here, would be grateful for some help. Thanks!

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If the question is: find x and y then this is how...

Real part:
2x-1=9
so x=4
Imaginary part:
5i=(y-1)i
so y=6

:)

Question 739124: (7 + i ) – (6 – 2i)
I know the answer is 1 + 3i but i dont know how to get there

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(7 + i ) – (6 – 2i)=(7-6)+(1-(-2))i=1+3i

Question 743488: in a right triangle, find the length of the not given. a=8, c=17

Question 742312: 2/i-3
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I think you are trying to say this:

so here goes...

Question 746369: Product of (1-2i)&(-1+2i)

Question 746708: graph the equation of x^2 + y^2 - 36
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What?

An equation must be stated as something is equal to something. A function would be given a name and then equated to an expression. The polynomial alone as given is not enough. Do you want the function, or do you want the quadratic expression equal to 0 or some constant?

The given expression appears to be for a circle. Is the expression equal to 0?

Best guess,

which would be a circle centered at the origin, with radius 6 units.

4/2+(1/x)

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4/2+(1/x)
-----------------
Equations are solved, not simplified.
But, this is not an equation - it has no equal sign.
4/2+(1/x)
= 2 + 1/x
=

Question 745609: I need help on how to use DE Moivre's Theorem for the following,
(3-3squareroot(3i))^7 ( to the seventh)
then put it in complex form and polar(express in radians)
i have been trying for days to solve it , so if you can help it will very greaatly appreciated. :) Thanks

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I need help on how to use DE Moivre's Theorem for the following,
(3-3squareroot(3i))^7 ( to the seventh)
then put it in complex form and polar(express in radians)
-------------
Is the i inside the radical?

Question 745496: simplify complex fraction: numerator is x/(x+2) and denominator is 1/x + 1/x+2. I think answer is x^2/2x+2 but not sure how to get it.
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simplify complex fraction: numerator is x/(x+2) and denominator is 1/x + 1/x+2
note that 1/x +1/(x+2) = x+2+x/x(x+2) then we get
x/(x+2) * x(x+2)/2x+2
cancel the x+2 term and we have
x * x / 2(x+1) = x^2 / 2(x+1)

Question 745050: Proof
a*(b*c)+b*(c*a)+c*(a*b)=0

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Proof
a*(b*c)+b*(c*a)+c*(a*b)=0
---------------
a*(b*c)+b*(c*a)+c*(a*b) = 3abc, not zero for all values of a, b & c.

Question 744694: I'm having some trouble with a question on my test:
.
-11x^2+12x - 6 = 0
.
.
A) x ={-6 +- i[SQRT(60)]}/11
B) x = {-6 +- i[SQRT(30)]}/-11
C) x = {6 +- i[SQRT(12)]}/6
D) x = {-6 +- i[SQRT(22)]}/-11
E) x = {-11 +- i} /6
F) x = {11 +- i} /-6
.
.
What I did first is to insert the question into the quadratic formula, with -11, 12 and -6 as a, b and c.
(-12 ± √144 - 264) / (-22)
Which turns into
(-12±√-120) / -22
.
But now I'm stuck. Could I have some help please?

You can put this solution on YOUR website!
-11x^2+12x - 6 = 0
.
.
A) x ={-6 +- i[SQRT(60)]}/11
B) x = {-6 +- i[SQRT(30)]}/-11
C) x = {6 +- i[SQRT(12)]}/6
D) x = {-6 +- i[SQRT(22)]}/-11
E) x = {-11 +- i} /6
F) x = {11 +- i} /-6
.
.
What I did first is to insert the question into the quadratic formula, with -11, 12 and -6 as a, b and c.
(-12 ± √144 - 264) / (-22)
Which turns into
(-12±√-120) / -22
.
But now I'm stuck. Could I have some help please?
-----------------------
(-12±√-120) / -22 You were almost done.
=
=
--> C
============
 Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . The discriminant -120 is less than zero. That means that there are no solutions among real numbers. If you are a student of advanced school algebra and are aware about imaginary numbers, read on. In the field of imaginary numbers, the square root of -120 is + or - . The solution is , or Here's your graph:

Question 744121: if i=squareroot of -1 and a and b are non-zero real numbers, what is 1/a+bi?
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if i=squareroot of -1 and a and b are non-zero real numbers,
what is 1/a+bi?
------
Multiply numerator and denominator by a-bi to get:
----
= (a-b)/(a^2+b^2)
==================
Cheers,
Stan H.
==================

Question 743912: how do i put(3 + 4i) + (8 + 2i) as a complex number in standard form..?
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how do i put(3 + 4i) + (8 + 2i) as a complex number in standard form..?
-------------

Question 743616: I need some help with this question
-4 1/5x2 + 1 2/5x - 2 4/5 = 0
Here are my options. I can't seem to get it right.
A) x = {1 +- i[SQRT(23)]}/6
B) x = {-1 +- i[SQRT(10)]} / 6
C) x = {1 +- [SQRT(23)]} /-6
D) x = 1 +- 5/2 x^2
E) x = -1 +- 2/5 x^2
F) x = 2 +- 20 x^2

Answer by Edwin McCravy(8920)   (Show Source):
You can put this solution on YOUR website!
x² + x -  = 0

Change the mixed fractions to improper fractions:

x² + x -  = 0

Clear of fractions by multiplying every term through by 5

-21x² + 7x - 14 = 0

Divide through by -7, (negative so the leading term will be
positive):

3x² - x + 2 = 0

That doesn't factor so we use the quadratic formula:

x =

with a = 3, b = -1, c = 2

x =

x =

x =

x =

x =

Edwin

Question 743569: how to work this problem: 4-2i / -3i
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how to work this problem: 4-2i / -3i
------
Multiply numerator and denominator by 3i
-----
= [3i(4-2i)]/9
-----
= [6 + 12i]/9
------
= (2/3) + (4/3)i
===================
Cheers,
Stan H.
===================

You can put this solution on YOUR website!

Next we multiply out both numerator and denominator:

Note that

and

So our expression becomes

Which simplifies to:

or

Question 742450: what is the solution for :(h-6)(h+8)
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use zero product rule

if =>
if =>

Question 742206: 4-i over
2-6i
can you walk me through the steps Im a little rusty.
This is what I got.
2 % i
7

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PROBLEM:
4-i/2-6i
4+-6.5i

Question 740834: Write the expression (3 − 9i) + (4 + 5i) as a complex number in standard form.
Found 2 solutions by lynnlo, solver91311:
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(3-9i)+(4+5i)
3+4============7
9i-5i==========4i
=====7-4i

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To add two complex numbers, add the real parts and then add the coefficients of the imaginary parts.

John

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it

Question 740092: I'm having trouble with a question in my lesson. Could I have some help?
(-1 + [SQRT(3)]i)^3
These are the options:
A) 2 + 2[SQRT(3)]i
B) -2 - 2[SQRT(3)]i
C) 0
D) -8
E) -8 + 8[SQRT(3)]i
F) 8

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You can calculate as you would calculate the cube of any binomial, using the formula

If you already learned about the polar form of complex numbers, you could transform the number into its polar form, if that made it easier for you.
has
absolute value (or modulus):
argument:
Multiplication and powers are easier in polar form.
To multiply, you just multiply the absolute values and add the arguments.
So

Question 737751: What is the jth root of j? Give the exact value.
Found 3 solutions by Ed Parker, Edwin McCravy, MathLover1:
Answer by Ed Parker(21)   (Show Source):
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I'm the same person as Edwin McCravy above.

Let's simplify

So we have

So we find

Now to continue we must find an expression for ln(j)

Euler's equation is

Substitute

Therefore since

then

Now we go back to :

Now we go back to

.

Edwin

Answer by Edwin McCravy(8920)   (Show Source):
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The other tutor's answer is wrong.

Apparently, you're an electronics student who uses i for current, and
therefore has to use j for √-1.

Sorry, I thought you wanted jj and worked that out.  I'll redo it
and put the results here later.  It's actually 4.810477381.  But
here is

Now to continue we must find an expression for ln(j)

Euler's equation is

Substitute

Therefore since

then

Now we go back to where we were:
= 0.2078795764

Edwin