Question

Find the real or imaginary solutions of each equation by factoring.
Did I do this right? And what's the difference between real and imaginary solutions? How do you tell the difference?
x^3 – 64 = 0
x(x^2-64)=0
x=0
(x+8)(x-8)=0
x+8=0
x-8=0
x=0,-8,8

Answer 1

64 is the cube of which number ?

Answer 2

if it was x^3 -64x , then you would have been correct!

Answer 3

Darn no it's not 64x
What do I do to figure this out?

Answer 4

use this formula
\(a^3-b^3 = (a-b)(a^2+ab+b^2)\)

Answer 5

here a = x
b =4

Answer 6

So
x^3-64^3=(x-64)(x^2+64x+64^2)?

Answer 7

the question is not 64^3
its 64, right ?

64 is the cube of what number?

Answer 8

4

Answer 9

yes
so its actually
x^3-4^3

Answer 10

now use the formula

Answer 11

x^3-4^3=(x-4)(x^2+4x+4^2)

Answer 12

yes
that = 0

so x-4 = 0 , x= ....

(x^2+4x+4^2) = 0
can you solve this quadratic ?

Answer 13

\[x=\frac{ \pm b \sqrt{b^2 -4ac} }{ 2 }\]
This formula?

Answer 14

yes
\(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 15

can you find a,b,c ?

Answer 16

Also x=4 for x-4=0
Yeah I thought I clicked the a after the 2 lol.

Answer 17

yes, x= 4 is correct :)

Answer 18

a=1
b=4
c=16

Answer 19

food!
now find b^2-4ac = ... ?

Answer 20

to answer your questions :
And what's the difference between real and imaginary solutions?
A : imaginary solutions have 'i' , imaginary unit \(i = \sqrt {-1}\) in them.
they occur when we get a negative value inside square root.

How do you tell the difference?

A: if b^2 -4ac = positive, solutions are real
if b^2-4ac is negative, then solutions are imaginary!

so, here, whats b^2-4ac ?

Answer 21

16-64=-48

Answer 22

yes! so it came out to be negative!
so your solutions will be complex :)

Answer 23

now use the formula
\(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 24

\[x=\frac{ 4 \sqrt{-48} }{ 2 }\]
\[x=\frac{ -4 \sqrt {-48} }{ 2 }\]

Answer 25

you forgot \(\pm\) sign in between

Answer 26

Indeed I did oops.

Answer 27

so, could you simplify that ?

Answer 28

Would it be
x=-2 +,- sqrt -48 ?

Answer 29

sqrt (-48) itself could be simplified...

can you factor 48 ?

Answer 30

No.

Answer 31

\(48 = 16\times 3 \\ \sqrt {-48} = \sqrt {16} \sqrt {-3} = 4 \sqrt {-3}\)
does that make sense ?

Answer 32

Ahh okay! I know how to do that waqsn't sure that's what you meant.

Answer 33

okk..
and \(4 \sqrt {-3} = 4 \sqrt {-1}\sqrt 3 = 4i \sqrt 3\)

Answer 34

Okay confused.
I don't know how to write this...

Answer 35

I have to show my work and I'm stuck.

Answer 36

i just used the fact that \(\sqrt{-1} = i\)
nothing else...

Answer 37

Yeah I'm stuck at the factoring part
after I have
\[x=\frac{ -4 \pm \sqrt {-48} }{ 2 }\]
What was my next step after that?

Answer 38

i already shown
\(\sqrt {-48} = 4i \sqrt 3\)

Answer 39

Yes I got that...
So what
\[x=-2 \pm 4i \sqrt{3}\]