*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.

- rebeccaxhawaii

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- rebeccaxhawaii

@johnweldon1993

- rebeccaxhawaii

okay so this is only the first question

- johnweldon1993

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- rebeccaxhawaii

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.

- johnweldon1993

...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

- johnweldon1993

So we just need to come up with some function
\(\large ax^2 + bx + c\) where \(\large b^2 - 4ac\) fits those criteria

- Owlcoffee

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.

- rebeccaxhawaii

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 ?

- johnweldon1993

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

- rebeccaxhawaii

wtf

- johnweldon1993

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

- rebeccaxhawaii

i thought g(x) doesnt have to be

- Owlcoffee

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

- johnweldon1993

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

- Owlcoffee

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.

- rebeccaxhawaii

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

- johnweldon1993

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

- rebeccaxhawaii

john john john .. can you talk rebecca style please

- johnweldon1993

XDDDD okay..lets see

- johnweldon1993

Let me give you an example, maybe that will help

- rebeccaxhawaii

okay okay

- johnweldon1993

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

- johnweldon1993

Tell me ANYTHING that doesn't make sense

- rebeccaxhawaii

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

- johnweldon1993

Lol I got time :)

- johnweldon1993

Actually...it might help reading about it elsewhere as well?
http://www.regentsprep.org/regents/math/algtrig/ate3/discriminant.htm

- rebeccaxhawaii

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

- johnweldon1993

Okay, sorry if I hurt your head :/

- rebeccaxhawaii

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

- rebeccaxhawaii

any who get good rest youll need it

- johnweldon1993

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

- rebeccaxhawaii

@johnweldon1993

- johnweldon1993

Oh back to this story!! Almost forgot it, gave me chills XD
Okay first, I want you to read that link I posted above :)

- rebeccaxhawaii

okay sorry im reading it

- johnweldon1993

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

- rebeccaxhawaii

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

- johnweldon1993

XDDDDD lol *foot in mouth* XDDD

- rebeccaxhawaii

okay i think i got it

- johnweldon1993

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

- rebeccaxhawaii

what is that

- johnweldon1993

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?

- rebeccaxhawaii

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

- johnweldon1993

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

- rebeccaxhawaii

REBECCA WORDS PLS

- johnweldon1993

Lol no these are JOHN WORDS!!!
But i mean, you should get that though, based on the link you read

- rebeccaxhawaii

expecting me to read johnny dont expect miracles

- rebeccaxhawaii

jk ima look again

- johnweldon1993

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

- rebeccaxhawaii

wo complex roots ?

- rebeccaxhawaii

two *

- johnweldon1993

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

- rebeccaxhawaii

IM SO CONFUSED

- johnweldon1993

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

- rebeccaxhawaii

this is my type of learning

- rebeccaxhawaii

okay 593

- johnweldon1993

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?

- rebeccaxhawaii

indeedy

- johnweldon1993

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?

- rebeccaxhawaii

yes

- johnweldon1993

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 :)

- rebeccaxhawaii

wait so how is it a perfect square ?

- rebeccaxhawaii

i understand everything else

- johnweldon1993

Tell me what a perfect square is :)

- rebeccaxhawaii

|dw:1446577027112:dw|

- johnweldon1993

XD you're perfect XD

- johnweldon1993

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

- rebeccaxhawaii

64

- johnweldon1993

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

- rebeccaxhawaii

yes

- johnweldon1993

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?

- rebeccaxhawaii

no its not

- johnweldon1993

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

- rebeccaxhawaii

okay yay i get it so far

- johnweldon1993

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

- rebeccaxhawaii

yes

- johnweldon1993

Okay good! that's 1 done......we have g(x) done
Now lets work on f(x)
So again...give me 3 random numbers

- rebeccaxhawaii

im gonna write this down

- johnweldon1993

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

- rebeccaxhawaii

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

- johnweldon1993

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

- rebeccaxhawaii

hah jk but my numbers are 274

- johnweldon1993

;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 :)

- rebeccaxhawaii

924

- johnweldon1993

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??

- rebeccaxhawaii

no i mean 294

- johnweldon1993

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??

- rebeccaxhawaii

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

- johnweldon1993

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

- rebeccaxhawaii

sweet as a lemon

- rebeccaxhawaii

so i win right ?

- johnweldon1993

Lol you win for f(x) yes...we found 3 numbers that make the discriminant a perfect square greater than 0
So \(\large f(x) = 2x^2 + 9x + 4\)

- johnweldon1993

Now we just need h(x)
And you have already done it actually...remember where I said "we're coming back to this one"

- rebeccaxhawaii

okok im ready

- johnweldon1993

So you gave me other numbers before this one
a = 9
b = 2
c = 4
That gave us \(\large 2^2 - 4(9)(4)\) = 4 - 144 = -140\)
From that link...we saw that if we get a negative number from this *like we have* that will result in 2 complex solutions

- rebeccaxhawaii

yay #1 done

- johnweldon1993

<3

- rebeccaxhawaii

okay #2

- rebeccaxhawaii

Like magic, the doors creak open with a burst of dust, musty air, and shrieking bats. You lead McMerlock down the first corridor. It is so dark that you can barely see your own out–stretched hand. The darkness consumes you as you continue to trek deeper and deeper in the hallway. Suddenly, a mystical suit of armor blocks your path. A voice emanates from inside it.
“To pass by me you must tell me how to convert standard form into the general, vertex form... I have a test on it next week.”
Explain how to convert f(x) into the general, vertex form of the equation. Use complete sentences. You may use the f(x) you created in question 1 as an example.

- johnweldon1993

Omg the story continues!!!!!!

- rebeccaxhawaii

ikik contain your excitement please

- johnweldon1993

XD dont tell me what to do!!!
Okay so f(x) was
\[\large f(x) = 2x^2 + 9x + 4\]
In order to turn this into vertex form *gonna be a little messy* we need to complete the square

- rebeccaxhawaii

CONTAIN IT
CONCEAL DONT FEEL

- rebeccaxhawaii

4 ?

- johnweldon1993

Why did you write 4? lol

- rebeccaxhawaii

thats the |dw:1446580246579:dw|

- johnweldon1993

Lol the f(x) is the perfect square equation
\[\large f(x) = 2x^2 + 9x + 4\]
This is what we'll be working with

- rebeccaxhawaii

ok ok so wdym by that being a perfect square

- johnweldon1993

Hey hey hey no! we already went through this
remember the \(\large ax^2 + bx + c\)
That was when a = 2, b = 9 and c = 4...and when we plugged it into the \(\large b^2 - 4ac\) and got the perfect square?

- rebeccaxhawaii

4 is the perfect square i thought

- johnweldon1993

Nope...that is just part of the equation
Forget the whole word perfect square for this part :)

- rebeccaxhawaii

easily done

- johnweldon1993

Okay...so this part...we are just using the
\[\large f(x) = 2x^2 + 9x + 4\]
To convert this to vertex form...we have to complete the square..any idea how to do that?
*dont you dare draw a square again you butt! ;P

- rebeccaxhawaii

|dw:1446581037555:dw| ?

- rebeccaxhawaii

jkjk wait im reading it

- rebeccaxhawaii

no i do not

- johnweldon1993

Lol I'm gonna kick your butt! XD
Okay so completing the square...this is gonna get SLIGHTLY messy...but I know you can do it

- rebeccaxhawaii

mhmmm go on

- johnweldon1993

So...we have
\[\large y = 2x^2 + 9x + 4\]
First thing we need to do is subtract 4 from both sides of this equation
\[\large y - 4 = 2x^2 + 9x\]
And now we need to divide everything by 2...because we need that x^2 term alone
So
\[\large \frac{y}{2} - 2 = x^2 + \frac{9}{2}x\]
*I'ts only messy because we have that fraction now...all good up to now?

- rebeccaxhawaii

*writing*

- rebeccaxhawaii

okay im good

- johnweldon1993

Okay so
We need to take the coefficient of the 'x' term *which is 9/2......we need to divide that by 2.....which is 9/4.....and the we need to square that result....so 81/16
And we add that to both sides of this equation
\[\large \frac{y}{2} - 2 + \frac{81}{16} = x^2 + \frac{9}{2}x + \frac{81}{16}\]
With me still? ask question as we go if you need it

- rebeccaxhawaii

wtf

- johnweldon1993

Lol i told you it was gonna get messy...but it's not that bad of a process

- johnweldon1993

It simplifies down nicer though

- rebeccaxhawaii

why 81 and 16 though

- johnweldon1993

What is 9/4 squared?
\[\large (\frac{9}{4})^2 = \frac{9^2}{4^2} = \frac{81}{16}\]

- johnweldon1993

Make sense? the logic not the process...the process is new of course you wouldnt get that perfectly yet

- rebeccaxhawaii

BUT WHY

- johnweldon1993

Why what?
Well this is the process for completing the square
1) Get the x^2 and x term on one side
2) Get x^2 by itself..no coefficient
3) take the coefficient of the 'x' divide by 2...and square the result
4) add that to both sides
5) we'll go from there...

- rebeccaxhawaii

can you change the rules please

- rebeccaxhawaii

but kay i get it at its best extent of my understanding

- rebeccaxhawaii

math cheating

- johnweldon1993

Lol hey, I'm desirable XDDDD jk jk hha
But okay...just stick with me until the end :)
We have
\[\large \frac{y}{2} - 2 + \frac{81}{16} = x^2 + \frac{9}{2}x + \frac{81}{16}\]
We can now write the right hand sides as a sum of squares
\[\large \frac{y}{2} + \frac{49}{16} = (x + \frac{9}{4})^2\]
Lol just understand the logic for now...get every math step I'm doing?

- rebeccaxhawaii

end ... of time until the world blows up

- rebeccaxhawaii

okay i got it wait why is it ^2

- johnweldon1993

\[\large (x + \frac{9}{4})^2 = (x + \frac{9}{4})(x + \frac{9}{4})= x^2 + 2(\frac{9}{4}x) + \frac{81}{16} = x^2 + \frac{9}{2}x + \frac{81}{16}\]
That's why its raised to the second power...i simplified it down

- rebeccaxhawaii

WHAT

- johnweldon1993

XD lol just notice for now that I wrote
\(\large x^2 + \frac{9}{2}x + \frac{81}{16}\) as \(\large (x + \frac{9}{4})^2 \)
You can look up how to complete the square and it will make sense then...but for now you got me :P

- rebeccaxhawaii

okay i got it

- johnweldon1993

You sure? Okay...
So we have
\[\large \frac{y}{2} + \frac{49}{16} = (x + \frac{9}{4})^2\]

- johnweldon1993

All we need to do now...is subtract 49/16 from both sides
\[\large \frac{y}{2} = (x + \frac{9}{4})^2 -\frac{49}{16}\]
And finally multiply everything by 2...to get right of y/2
\[\large y = 2(x + \frac{9}{4})^2 - \frac{49}{8}\]

- johnweldon1993

*Rid of..not right of*

- rebeccaxhawaii

is that it

- johnweldon1993

mmhmm that's it :)

- rebeccaxhawaii

yay

- rebeccaxhawaii

Satisfied with his notes, the guardian lets you pass into the next chamber. As you enter the dimly lit room, McMerlock points out that the floor tiles have numbers on them, and that some floor tiles are missing. On the ceiling is painted a cryptic message.
“Only the solutions of g(x) will lead you safely across.”
Find the solutions of g(x). Show each step.
You can feel the treasure of the Lost Island of Laplaya in your grasp, as you deftly step on the correctly numbered stones. The last test of knowledge lay before you and Professor McMerlock. A wise old woman sits in front of the treasure. You can hear the crackle of magic coursing through her finger tips. She asks you “Is completing the square a good method for solving when the Discriminant is negative?” Professor McMerlock stammers, but you are confident in your answer.
Justify if completing the square is a good method for solving when the Discriminant is negative. Use any of your three functions as an example and respond in complete sentences.
Satisfied with your answer, the wise old woman grants you the treasure of the Lost Island of Laplaya! The treasure is an awesome ability to complete the square! Way to go! Professor McMerlock is thankful he picked you and promises to call on you again for another adventure.

- rebeccaxhawaii

last 2 questions

- johnweldon1993

Okay so the first part
We just need to solve for g(x)
what was the function we chose for g(x) again?

- rebeccaxhawaii

5x^2+9x+3

- johnweldon1993

Okay...so we basically need to find when
\[\large 5x^2 + 9x + 3 = 0\]
And lets just do it the easy way with the quadratic formula
\[\large \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[\large \frac{-9 \pm \sqrt{9^2 - 4(5)(3)}}{2(5)}\]
\[\large \frac{-9 \pm \sqrt{81 - 60}}{10}\]
\[\large \frac{-9 \pm \sqrt{21}}{10}\]
Now just solve that
\(\large \frac{-9 + \sqrt{21}}{10}\) and \(\large \frac{-9 -\sqrt{21}}{10}\)

- rebeccaxhawaii

woah

- johnweldon1993

Mmhmm, buts its easy, just plug that into a calculator and you're good :D

- rebeccaxhawaii

okay so those are the soluutions why are there 2

- johnweldon1993

If we were to graph this function...we would have a parabola that passes through the x-axis at 2 different points...hence 2 solutions

- rebeccaxhawaii

bare with me slowly

- johnweldon1993

I'm here love :D

- rebeccaxhawaii

wdym now just solve that

- johnweldon1993

Just put it in a calculator
You have 2 different solutions so put
\[\large \frac{-9 + \sqrt{21}}{10}\]
and get that value and then do the other one too

- rebeccaxhawaii

okay LAST ONE HOW EXCITING

- rebeccaxhawaii

HELLO

- johnweldon1993

I mean...the last one shouldn't technically be a question...more of an opinion
But yes it is a good method
Lets use h(x) as an example
\[\large 9x^2 + 2x + 4\]
Lets complete the square *gonna get really messy so I'll do it*
\[\large 0 = 9x^2 + 2x + 4\]
Subtract 4 from both sides
\[\large - 4 = 9x^2 + 2x\]
divide everything by 9
\[\large - \frac{4}{9} = x^2 + \frac{2}{9}x\]
Take the coefficient of 'x' ...divide by 2...square the result, add to both sides
\[\large - \frac{4}{9} + \frac{4}{324} = x^2 + \frac{2}{9}x + \frac{4}{324}\]
Write the right hand side as sum of squares
\[\large -\frac{35}{81} = (x + \frac{2}{18})^2\]
Square root both sides
\[\large \sqrt{-\frac{35}{81}} = x + \frac{2}{18}\]
We have an imaginary number! *square root a negative number*
\[\large \frac{\sqrt{35}}{9}i = x + \frac{2}{18}\]
Finally subtract 2/18 from both sides
\[\large x = \pm\frac{\sqrt{35}}{9}i - \frac{2}{18}\]

- johnweldon1993

Sorry a lot to type! lol

- rebeccaxhawaii

omg no you are literally a saint taking the time to educate the uneducated gosh you are an angel thank you

- johnweldon1993

Lol I got you :D

- johnweldon1993

Does everything make sense *for the most part anyways* Is Professor McMerlock happy? ;D

- rebeccaxhawaii

haha yes i am thank you so much until next time

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