Question

Medal and fan. What are the real or imaginary solutions to this equation x^4-52x^2+576=0

Answer 1

\[x^4-52x^2+576=0\] does it factor?

Answer 2

Again (i'm bad at factoring) I have to find something that adds to -52 and multiplies to 576?

Answer 3

yeah

Answer 4

try \(16\) and \(36\)

Answer 5

So -16 and -36, How you came up with that that quickly I will never know.

Answer 6

i cheated
i suck at factoring
but we can do this without factoring if you like
might even be easier

Answer 7

Eh, I should probably learn to factor if that's alright?

Answer 8

Or at least practice it.

Answer 9

well you should really learn how to solve a quadratic equation without factoring
most of them do not factor using whole numbers in any case

Answer 10

Using the quadratic formula yeah?

Answer 11

and how you would know to come up with 16 and 36 is anyone’s guess
i would not use the quadratic formula for this one
i would complete the square, much easier here

Answer 12

Okay, there's another way lol. Teach me how to complete the square? /: I'll be your friend forever.

Answer 13

ok lets start with an easy one then we do this one ok?

Answer 14

Sounds good.

Answer 15

lets start with \[x^2+6x-5=0\]
add \(5\) to both sides, what do you get?

Answer 16

Just double checking, It will give me both answers right? I know this calls for two.

Answer 17

x^2+6x=5

Answer 18

yeah we will finish this in a moment
ok now you have
\[x^2+6x=5\] what is half of \(6\) ?

Answer 19

3.

Answer 20

right
what is \(3^2\)?

Answer 21

9

Answer 22

ok good so next step is to turn
\[x^2+6x=5\] in to
\[(x+3)^2=5+9\\
(x+3)^2=14\]

Answer 23

Yay. Good luck me.

Answer 24

a word of explanation
\[(x+3)^2=x^2+6x+9\] so by adding 9 to both sides you changed
\[x^2+6x=5\] to
\[(x+3)^2=14\]

Answer 25

that makes
\[x+3=\pm\sqrt{14}\] and so
\[x=-3\pm\sqrt{14}\]

Answer 26

notice you could never in a million years factor
\[x^2+6x-5=0\]

Answer 27

Okay so. Now I need to know the sqrt of 14? And solve

Answer 28

Yep.

Answer 29

yes that is right
when you take the square root, don't forge the \(\pm\)

Answer 30

So 3 +- 3.7

Answer 31

So 6.7 and -.7?

Answer 32

leave it with the radical
\[3\pm\sqrt{14}\] looks nicer
forget the decimals

Answer 33

but yeah you are right

Answer 34

Okay, But on the next I have to solve it out.

Answer 35

shall we finish this problem? the one you posted?

Answer 36

Yes, please.

Answer 37

well that was wrong in any case

Answer 38

do you want to believe it factors as
\[(x^2-36)(x^2-16)=0\] or do you want to complete the square again?

Answer 39

Lol! I was about to feel incredibly smart for finding a typo! It factors as (x-16)(x-36) yeah?

Answer 40

Can we solve this way?

Answer 41

yeah a typo
but you are probably incredibly smart in any case

Answer 42

psst...difference of two squares ;-)

Answer 43

Wellllllllll, thanks for that (:

Answer 44

the factor way is easy if we know if factors
both \(x^2-36\) and \(x^2-16\) are the difference of two squares

Answer 45

so factor all the way down and you get the zeros, more or less in your head

Answer 46

The difference of two squares? I'm confused. Don't lose me just yet.

Answer 47

\[a^2-b^2=(a+b)(a-b)\]
\[x^2-36=(x+6)(x-6)\]

Answer 48

similar for \(x^2-16\)

Answer 49

$$\Huge a^2-b^2=(a+b)(a-b)$$

Answer 50

So that's the equation to find the difference of two squares?

Answer 51

yeah i always factors that way
of course if you had
\[x^2-16=0\] and you were a thinking human being you would write
\[x^2=16\\x=4,x=-4\]

Answer 52

Oh wait! okay I see that. So then the equation for x^2-16 would be x^2-16=(x+4)(x-4)

Answer 53

yes exactly

Answer 54

Ha! I'm a thinking human then. So are those my answers? Though How do I know which two?

Answer 55

not two , four!

Answer 56

So they're all four my answer?

Answer 57

Okay. I see that. Can I get your help on one more (it's similiar)? /:

Answer 58

\[x^4-52x^2+576=0\] a polynomial of degree 4
this one has 4 real zeros
once you have it factored all the way down to its bare bones it is
\[(x+6)(x-6)(x+4)(x-4)=0\]

Answer 59

as you can see this problem has been cooked to within an inch of its life to give such nice answers
mostly they don't work out so nicely

Answer 60

sure go ahead, we can do another one

Answer 61

Okay. This one says the same, to find the real and imaginary answers of x^3=216

Answer 62

$$\Huge ab=0 \iff a=o \text{ or } b=0.$$

Answer 63

this one is different, you want to start with
\[x^2-216\] and factor as the difference of two cubes

Answer 64

did you mean x^3? Or..

Answer 65

of course in order to do that you have to
a) recognize that \(216=6^3\) and that
b) \(a^3-b^3=(a-b)(a^2+ab+b^2)\)

Answer 66

Alright, this is tough. Walk me through it and I'll do the math. Just point the way.

Answer 67

yeah i meant
\[\huge x^{\color{red}3}-216=0\]

Answer 68

\[x^3-6^3=0\] is how you visualize it

Answer 69

then use
\[a^3-b^3=(a-b)(a^2+ab+b^2)\] with
\[a=x,b=6\]

Answer 70

Okay. So. \[x^3-6^3=(x-6)(x^2+6x+6^2) \]

Answer 71

yes aka
\[(x-6)(x^2+6x+36)=0\]

Answer 72

Okay. You're very fast at typing equations, if only there was a medal. (;

Answer 73

now one answer is obvious
\[x-6=0\\
x=6\] the other two complex zeros you get are from solving
\[x^2+6x+36=0\] the solutions are not real
you want to find them?

Answer 74

I can obviously try lol. Lets see..

Answer 75

\[x^2+6x=-3\]

Answer 76

-36**

Answer 77

\[x^2+x=-6\]

Answer 78

Am I doing this right. Because I really don't think I am.

Answer 79

ok good
then \[x^2+6x=-36\]is the first step
next step
what is half of 6?

Answer 80

3.

Answer 81

and \(3^2\)>

Answer 82

9?

Answer 83

no, not 9?, but 9!
so we go from
\[x^2+6x=-36\] to \[(x+3)^2=-36+9\\
(x+3)^2=-27\]

Answer 84

like i said, the answers are complex because now when you take the square root you will be taking the square root of a negative number

Answer 85

Okay. Yeah, imaginary numbers.

Answer 86

you get
\[x+3=\sqrt{-27}\] you know how to write that as a complex number?

Answer 87

actually you get
\[x+3=\pm\sqrt{-27}\]

Answer 88

Uhh lol would it be 3+ something I and 3-something i

Answer 89

actually since \(27=9\times 3\) it is
\[\sqrt{-27}=3\sqrt{3}i\]

Answer 90

making your final answer
\[-3\pm3\sqrt{3}i\]

Answer 91

Alright. I think I understand heh. Thank you so much!

Answer 92

yw