Question

To open these doors, you must speak three functions in standard form.
One function, f(x), with two real rational solutions.
One function, g(x), with two real irrational solutions.
One function, h(x), with two complex solutions.

Answer 1

it is Completing the square

Answer 2

Weird doors.
But anyway, are those the only requirements?

Answer 3

yess

Answer 4

Ok, well, what are the two possible values for \(\sqrt{4}\)?

Answer 5

2 and -2?

Answer 6

Yes. So an equation like \(x^2 = 4\) would satisfy \(f(x)\). Or since it has to equal 0 to solve it, \(f(x) = x^2 - 4\)

Answer 7

ohh! okay:) thankss

Answer 8

You probably get the method and can do the other two now. :D

Answer 9

yes for sure!:) thanks mann

Answer 10

@SACAPUNTAS , hey one more question.. what is an irrational solution?

Answer 11

A solution where the number is irrational. Do you know what an irrational number is?

Answer 12

no:(( sorry

Answer 13

I don't know the exact formal definition. But it's basically any number you can't write as a nice fraction or decimal.

Answer 14

okay so would it be okay if you explained the second equation to me? i just wanna know so i can have a good grade in math, ya know?

Answer 15

Well, this one is actually rather easy. Think about it! Most square roots are hideously ugly numbers.

Answer 16

For example, the solutions to \(x^2 =2\) cannot be written out in any way that looks nice.

Answer 17

(Except, of course, as \(x = \pm\sqrt{2}\) but you see what I mean.

Answer 18

yeah! so an equation i could use would be like g(x)=?

Answer 19

Look at \(f(x)\) up there and you tell me! It's the same form. :D

Answer 20

g(x)= 2x^2?

Answer 21

Why did you multiply. Up there you subtracted.

Answer 22

idk, i am so confused.

Answer 23

Ok. Want me to try explaining it again?

Answer 24

i get x^2=0, but to have it setup with g(x)

Answer 25

x^2=2 i ment..

Answer 26

That's the part I'll explain!

Answer 27

then yes explain away!:)

Answer 28

Ok, the solutions to an equation are the points where y (or f(x) or g(x) or whatever = 0)

Answer 29

There are sophisticated mathematical reasons for this but I can't articulate them so let's just say that's the case and go with it.

Answer 30

yes!

Answer 31

In other words, if we want to stick f(x) or g(x) or whatever in there, we have to replace a 0.

Answer 32

How can we make one side 0 when \(x^2 = 2\)?

Answer 33

subtract 2!!:) your a genius!! okay so g(x)=x^2-2?

Answer 34

Yes!

Answer 35

i wish i could give you a nether metal!

Answer 36

I wish you could too. :P

Answer 37

:) hah

Answer 38

Anyway... do you get complex numbers?

Answer 39

kind of... but can i just go over the next 2 to make sure that i got it?

Answer 40

Sure.

Answer 41

1. f(x)=x^2-4
2. g(x)=x^2-2
3. wait jk i dont get that. wow i feel stupid

Answer 42

I'm happy to help. I don't want to just give you the answers though.
(You felt better when you figured it out yourself, didn't you? :D )

Answer 43

yes for sure!! just give me little crumbss

Answer 44

Ok, so you get f(x) and g(x), right?

Answer 45

yes!

Answer 46

Ok, so a complex number means a number that can be written in form \(a + bi\) where i is the imaginary number.

Answer 47

okay so i could just plug in any number for a and b?

Answer 48

example. 2+4i=h(x)

Answer 49

That's certainly valid, but I have a feeling they want something in terms of x.

Answer 50

okay. so h(x)=2x+4i?

Answer 51

That only has one complex solution.

Answer 52

Do you know what the definition of i is?

Answer 53

it is an imaginary number.. i= negative 1 in a square root. i^2= -1..is that what u mean?

Answer 54

Right. So to get a complex number we just need an equation where we're trying to take the square root of something negative.

Answer 55

okay.. So h(x)=-3x+11i?

Answer 56

There are no squares in there! Like, look at the ones we did already.
To solve \(x^2 - 4\) we took the square root of 4.
To solve \(x^2 - 2\) we took the square root of 2.

Answer 57

h(x)=-3x^2+11i?

Answer 58

You know, the way the question is worded, that's actually a perfectly valid answer.

Answer 59

thank gooodness! i thought i would never get it!:) thanks so much for your help

Answer 60

Well, I don't think it's actually what they're going for... but it's also not wrong. So I'm a little torn on what to say. :P

Answer 61

okay im going with it!!!:)

Answer 62

Ok. If you get marked wrong, point out that the answer simply asked for two complex solutions, and did not specify the function had to contain only real numbers.

Answer 63

okay, i really cannot thank you enough!!:D