Question

*boring story that tries to make math interesting*

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

@johnweldon1993

Answer 2

okay so this is only the first question

Answer 3

....I wanna know the story >.< XD

Answer 4

if you must ...


For this adventure, you and world renowned Professor Sherlock McMerlock are traveling to the Lost Island of Laplaya. When the boat arrives on the island shore, you and Professor McMerlock disembark on your adventure. You trudge through the jungles and arrive at three impressively large doors. About eye level on each door is an intricately carved keyhole. Directions are scratched into the wood above each keyhole.

Answer 5

...It's SO GOOD!!!

Lol okay..so

It relates to the quadratic formula, more specifically the discriminant

\[\large \sqrt{b^2 - 4ac}\]

If \(\large b^2 - 4ac\) = a perfect square...we have 2 rational number solutions
if \(\large b^2 - 4ac\) does not = a perfect square, we have 2 irrational number solutions
and if \(\large b^2 - 4ac\) is less than 0...we get 2 complex solutions

Answer 6

So we just need to come up with some function

\(\large ax^2 + bx + c\) where \(\large b^2 - 4ac\) fits those criteria

Answer 7

You can define any type of function with a standard form, let it be polynomial or exponential.
For example, the first function can be represented as a polynomial in order to keep it simple, remember that rational are part of the real numbers, rational.
So we can plot any rational number on the structure: \[f(x)=(x-a)(x-a_2)...(x-a_n)\]
where all those "a"s represent the roots or the "solutions" to the function.
I'll take, for instance two numbers: \(\frac{ 5 }{ 4 }\) and \(-\frac{ 10 }{ 7 }\) which adds some variation:
\[f(x)=(x-\frac{ 5 }{ 4 })(x-(-\frac{ 10 }{ 7 })) \iff f(x)=(x-\frac{ 5 }{ 4 })(x + \frac{ 10 }{ 7 })\]
You can apply the distributive if you want to find the general form of that function.

Answer 8

so i found some answers would f(x)= x^2+6x+16 , g(x)= x^2+6x+1 , h(x)+2x^2+4x+5
would this work ?

Answer 9

f(x) \(\large 6^2 - 4(1)(16) = 36 - 72 = -36\) Nope...to be real...it has to be > 0

Answer 10

wtf

Answer 11

f(x) and g(x) both have to be real...so those discriminantes have to be > 0

Answer 12

i thought g(x) doesnt have to be

Answer 13

But what assures the solutions will be rational?
They will for sure be real, but consider that the solution might as well be irrational.

Answer 14

One function, f(x), with two ****real**** rational solutions. > 0
One function, g(x), with two ****real**** irrational solutions. > 0
One function, h(x), with two ****complex**** solutions. < 0

Answer 15

Yes, with that they mean a real number structured as a rational and irrational, else she would just put a "2" and it would solve the problem.

Answer 16

ok i gotcha but what do you mean by less than 0

Answer 17

The quadratic formula is \[\large \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Remember how I said the discriminant is that \(\large b^2 - 4ac\) part?

Well see how it is under the square root sign? if the discriminant comes out negative...we take the square root of a negative number....hence it would be complex solutions

Answer 18

john john john .. can you talk rebecca style please

Answer 19

XDDDD okay..lets see

Answer 20

Let me give you an example, maybe that will help

Answer 21

okay okay

Answer 22

So g(x) has to be 2 real irrational solutions

\[\large ax^2 + bx + c\]

I need some a,b and c that make \(\large b^2 - 4ac\) a number > 0

so hmm...lets do b = 6 , a = 1 and c = 2 *just completely random numbers* let see if it works?

\[\large b^2 - 4ac \rightarrow 6^2 - 4(1)(2) \rightarrow 36 - 8 \rightarrow 28\]

Since that is a number > 0 and we know 28 is NOT a perfect square

\(\large x^2 + 6x + 2\) Will have 2 real irrational numbers

Answer 23

Tell me ANYTHING that doesn't make sense

Answer 24

this might take a while to process this as you know im quite slow

Answer 25

Lol I got time :)

Answer 26

Actually...it might help reading about it elsewhere as well?

http://www.regentsprep.org/regents/math/algtrig/ate3/discriminant.htm

Answer 27

john im done with school in just a bit do you think we can continue tomorrow

Answer 28

Okay, sorry if I hurt your head :/ </3 XD

Answer 29

you didnt its the math thats killing me ... slowly ... like my brain

Answer 30

any who get good rest youll need it

Answer 31

Lol take a breather...get a snack...and some juice...but share :)

Answer 32

@johnweldon1993

Answer 33

Oh back to this story!! Almost forgot it, gave me chills XD

Okay first, I want you to read that link I posted above :)

Answer 34

okay sorry im reading it

Answer 35

Lol it's okay...math cheating on me it's cool ;P

Answer 36

actually the class is having time of prayer so yeah i guess they are similar lol

Answer 37

XDDDDD lol *foot in mouth* XDDD

Answer 38

okay i think i got it

Answer 39

Okay...so

I want a quadratic with 2 real IRRATIONAL solutions...

Answer 40

what is that

Answer 41

Lol your original question

"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."

So we need those...based on what you read...how can we do the g(x) one?

Answer 42

when we plug it in and check it it will work but idk how to get a problem like that

Answer 43

Okay...so based on what you read...if we have some function

\(\large ax^2 + bx + c\) where the discriminant is \(\large b^2 - 4ac\)

If the discriminant is greater than 0 and not a perfect square...we will have our g(x) satisfied

So we just pick random numbers for a,b and c until we make that work

Answer 44

REBECCA WORDS PLS

Answer 45

Lol no these are JOHN WORDS!!!

But i mean, you should get that though, based on the link you read

Answer 46

expecting me to read johnny dont expect miracles

Answer 47

jk ima look again

Answer 48

Remember my example I did before

I chose random numbers for a,b and c where after calculating \(\large b^2 - 4ac\) I got a number greater than 0 that wasnt a perfect square...so it would have worked for g(x)

I'll repost it

So g(x) has to be 2 real irrational solutions

\[\large ax^2+bx+c\]

I need some a,b and c that make \(\large b^2−4ac\) a number > 0

so hmm...lets do b = 6 , a = 1 and c = 2 *just completely random numbers* let see if it works?

\[\large b^2−4ac→6^2−4(1)(2)→36−8→28\]

Since that is a number > 0 and we know 28 is NOT a perfect square

\(\large x^2+6x+2\) Will have 2 real irrational numbers

Answer 49

wo complex roots ?

Answer 50

two *

Answer 51

Are you saying you want to do that next? Because that is not what I just did above...that was for the 2 REAL irrational roots

Answer 52

IM SO CONFUSED

Answer 53

Okay forget everything so far

Focus on this equation \(\large b^2 - 4ac\)

Give me 3 random numbers that I can plug in for a,b and c

Answer 54

this is my type of learning

Answer 55

okay 593

Answer 56

Lol I'm gonna get you through this hun :P

Okay so you're giving me 5 for a 9 for b and 3 for c right?

Answer 57

indeedy

Answer 58

Okay

So with a = 5, b = 9 and c = 3

From this equation \(\large b^2 - 4ac\) that would give me

\[\large 9^2 - 4(5)(3) = 81 - 60 = 21\]

Good so far?

Answer 59

yes

Answer 60

Okay...now comes the learning part

When \(\large b^2 - 4ac\) turns out to be greater than 0...and not a perfect square, then we would get 2 real irrational roots

Dont worry about why for now....just tell me if you accept that :)

Answer 61

wait so how is it a perfect square ?

Answer 62

i understand everything else

Answer 63

Tell me what a perfect square is :)

Answer 64

XD you're perfect XD

Answer 65

Okay...but should i take that to mean you're not sure what a perfect square number is?

Answer 66

64

Answer 67

Okay good...and the reason is because we know 8^2 = 64 right?

Answer 68

yes

Answer 69

Perfect, we're getting there :)

So in the numbers you gave me...and I went through the equation and got 21 as an answer...is 21 a perfect square?

Answer 70

no its not

Answer 71

Right!!!

So, we went through the equation...and got a number GREATER than 0....AND it was NOT a perfect square....from that link I gave you....that means we will have 2 real irrational roots

Answer 72

okay yay i get it so far

Answer 73

Okay good...lets finish that one up

You gave me a = 5, b = 9 and c = 3

Now we just need to plug those into the equation for the actual function...because that \(\large b^2 - 4ac\) was just for the first part...now the second part is using the equation

\(\large ax^2 + bx + c\) we just need to pug in the a,b and c you gave me

So the function would be

\(\large 5x^2 + 9x + 3\)

Answer 74

yes

Answer 75

Okay good! that's 1 done......we have g(x) done

Now lets work on f(x)

So again...give me 3 random numbers

Answer 76

im gonna write this down

Answer 77

Okay, I'll hang on for a bit :)

Answer 78

so what after a bit you just gonna leave okay i see how it is

Answer 79

Lol haha I was giving you time to write it down you butt!! ;P

Answer 80

hah jk but my numbers are 274

Answer 81

;P

Okay so lets try it out

a=2
b=7
c=4

So now...after we go through the \(\large b^2 - 4ac\) we WANT a number greater than 0...but we also want a perfect square this time...because that will give us 2 real rational solutions like we want for fx)

So lets try it

\(\large 7^2 - 4(2)(4) = 49 - 32 = 17\) sadly not a perfect square...so we need new numbers :)

Answer 82

924

Answer 83

Okay
a=9
b=2
c=4
\[\large 2^2 - 4(9)(4) = 4 - 144 = -140\]


Oh oh...okay wait forget f(x) for a sec...see how we came up with a negative number??

Answer 84

no i mean 294

Answer 85

Lol okay fine...we'll come back to the above though....

Okay so

a = 2
b = 9
c = 4

\[\large 9^2 - 4(2)(4) = 81 - 32 = 49\]
and is 49 a perfect square??

Answer 86

hold you applause everyone i did that to make it like that thank you thank you

Answer 87

Hahahahahaha I almost missed that you replied because i was still clapping ;)

Answer 88

sweet as a lemon

Answer 89

so i win right ?

Answer 90

Lol you win for f(x) yes...we found 3 numbers that make the discriminant a perfect square greater than 0

So \(\large f(x) = 2x^2 + 9x + 4\)

Answer 91

Now we just need h(x)

And you have already done it actually...remember where I said "we're coming back to this one"