rebeccaxhawaii
  • rebeccaxhawaii
*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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
rebeccaxhawaii
  • rebeccaxhawaii
@johnweldon1993
rebeccaxhawaii
  • rebeccaxhawaii
okay so this is only the first question
johnweldon1993
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • johnweldon1993
f(x) \(\large 6^2 - 4(1)(16) = 36 - 72 = -36\) Nope...to be real...it has to be > 0
rebeccaxhawaii
  • rebeccaxhawaii
wtf
johnweldon1993
  • johnweldon1993
f(x) and g(x) both have to be real...so those discriminantes have to be > 0
rebeccaxhawaii
  • rebeccaxhawaii
i thought g(x) doesnt have to be
Owlcoffee
  • 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
  • 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
  • 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
  • rebeccaxhawaii
ok i gotcha but what do you mean by less than 0
johnweldon1993
  • 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
  • rebeccaxhawaii
john john john .. can you talk rebecca style please
johnweldon1993
  • johnweldon1993
XDDDD okay..lets see
johnweldon1993
  • johnweldon1993
Let me give you an example, maybe that will help
rebeccaxhawaii
  • rebeccaxhawaii
okay okay
johnweldon1993
  • 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
  • johnweldon1993
Tell me ANYTHING that doesn't make sense
rebeccaxhawaii
  • rebeccaxhawaii
this might take a while to process this as you know im quite slow
johnweldon1993
  • johnweldon1993
Lol I got time :)
johnweldon1993
  • johnweldon1993
Actually...it might help reading about it elsewhere as well? http://www.regentsprep.org/regents/math/algtrig/ate3/discriminant.htm
rebeccaxhawaii
  • rebeccaxhawaii
john im done with school in just a bit do you think we can continue tomorrow
johnweldon1993
  • johnweldon1993
Okay, sorry if I hurt your head :/
rebeccaxhawaii
  • rebeccaxhawaii
you didnt its the math thats killing me ... slowly ... like my brain
rebeccaxhawaii
  • rebeccaxhawaii
any who get good rest youll need it
johnweldon1993
  • johnweldon1993
Lol take a breather...get a snack...and some juice...but share :)
rebeccaxhawaii
  • rebeccaxhawaii
@johnweldon1993
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
  • rebeccaxhawaii
okay sorry im reading it
johnweldon1993
  • johnweldon1993
Lol it's okay...math cheating on me it's cool ;P
rebeccaxhawaii
  • rebeccaxhawaii
actually the class is having time of prayer so yeah i guess they are similar lol
johnweldon1993
  • johnweldon1993
XDDDDD lol *foot in mouth* XDDD
rebeccaxhawaii
  • rebeccaxhawaii
okay i think i got it
johnweldon1993
  • johnweldon1993
Okay...so I want a quadratic with 2 real IRRATIONAL solutions...
rebeccaxhawaii
  • rebeccaxhawaii
what is that
johnweldon1993
  • 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
  • rebeccaxhawaii
when we plug it in and check it it will work but idk how to get a problem like that
johnweldon1993
  • 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
  • rebeccaxhawaii
REBECCA WORDS PLS
johnweldon1993
  • johnweldon1993
Lol no these are JOHN WORDS!!! But i mean, you should get that though, based on the link you read
rebeccaxhawaii
  • rebeccaxhawaii
expecting me to read johnny dont expect miracles
rebeccaxhawaii
  • rebeccaxhawaii
jk ima look again
johnweldon1993
  • 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
  • rebeccaxhawaii
wo complex roots ?
rebeccaxhawaii
  • rebeccaxhawaii
two *
johnweldon1993
  • 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
  • rebeccaxhawaii
IM SO CONFUSED
johnweldon1993
  • 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
  • rebeccaxhawaii
this is my type of learning
rebeccaxhawaii
  • rebeccaxhawaii
okay 593
johnweldon1993
  • 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
  • rebeccaxhawaii
indeedy
johnweldon1993
  • 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
  • rebeccaxhawaii
yes
johnweldon1993
  • 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
  • rebeccaxhawaii
wait so how is it a perfect square ?
rebeccaxhawaii
  • rebeccaxhawaii
i understand everything else
johnweldon1993
  • johnweldon1993
Tell me what a perfect square is :)
rebeccaxhawaii
  • rebeccaxhawaii
|dw:1446577027112:dw|
johnweldon1993
  • johnweldon1993
XD you're perfect XD
johnweldon1993
  • johnweldon1993
Okay...but should i take that to mean you're not sure what a perfect square number is?
rebeccaxhawaii
  • rebeccaxhawaii
64
johnweldon1993
  • johnweldon1993
Okay good...and the reason is because we know 8^2 = 64 right?
rebeccaxhawaii
  • rebeccaxhawaii
yes
johnweldon1993
  • 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
  • rebeccaxhawaii
no its not
johnweldon1993
  • 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
  • rebeccaxhawaii
okay yay i get it so far
johnweldon1993
  • 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
  • rebeccaxhawaii
yes
johnweldon1993
  • 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
  • rebeccaxhawaii
im gonna write this down
johnweldon1993
  • johnweldon1993
Okay, I'll hang on for a bit :)
rebeccaxhawaii
  • rebeccaxhawaii
so what after a bit you just gonna leave okay i see how it is
johnweldon1993
  • johnweldon1993
Lol haha I was giving you time to write it down you butt!! ;P
rebeccaxhawaii
  • rebeccaxhawaii
hah jk but my numbers are 274
johnweldon1993
  • 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
  • rebeccaxhawaii
924
johnweldon1993
  • 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
  • rebeccaxhawaii
no i mean 294
johnweldon1993
  • 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
  • rebeccaxhawaii
hold you applause everyone i did that to make it like that thank you thank you
johnweldon1993
  • johnweldon1993
Hahahahahaha I almost missed that you replied because i was still clapping ;)
rebeccaxhawaii
  • rebeccaxhawaii
sweet as a lemon
rebeccaxhawaii
  • rebeccaxhawaii
so i win right ?
johnweldon1993
  • 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
  • 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
  • rebeccaxhawaii
okok im ready
johnweldon1993
  • 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
  • rebeccaxhawaii
yay #1 done
johnweldon1993
  • johnweldon1993
<3
rebeccaxhawaii
  • rebeccaxhawaii
okay #2
rebeccaxhawaii
  • 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
  • johnweldon1993
Omg the story continues!!!!!!
rebeccaxhawaii
  • rebeccaxhawaii
ikik contain your excitement please
johnweldon1993
  • 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
  • rebeccaxhawaii
CONTAIN IT CONCEAL DONT FEEL
rebeccaxhawaii
  • rebeccaxhawaii
4 ?
johnweldon1993
  • johnweldon1993
Why did you write 4? lol
rebeccaxhawaii
  • rebeccaxhawaii
thats the |dw:1446580246579:dw|
johnweldon1993
  • 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
  • rebeccaxhawaii
ok ok so wdym by that being a perfect square
johnweldon1993
  • 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
  • rebeccaxhawaii
4 is the perfect square i thought
johnweldon1993
  • johnweldon1993
Nope...that is just part of the equation Forget the whole word perfect square for this part :)
rebeccaxhawaii
  • rebeccaxhawaii
easily done
johnweldon1993
  • 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
  • rebeccaxhawaii
|dw:1446581037555:dw| ?
rebeccaxhawaii
  • rebeccaxhawaii
jkjk wait im reading it
rebeccaxhawaii
  • rebeccaxhawaii
no i do not
johnweldon1993
  • 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
  • rebeccaxhawaii
mhmmm go on
johnweldon1993
  • 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
  • rebeccaxhawaii
*writing*
rebeccaxhawaii
  • rebeccaxhawaii
okay im good
johnweldon1993
  • 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
  • rebeccaxhawaii
wtf
johnweldon1993
  • johnweldon1993
Lol i told you it was gonna get messy...but it's not that bad of a process
johnweldon1993
  • johnweldon1993
It simplifies down nicer though
rebeccaxhawaii
  • rebeccaxhawaii
why 81 and 16 though
johnweldon1993
  • johnweldon1993
What is 9/4 squared? \[\large (\frac{9}{4})^2 = \frac{9^2}{4^2} = \frac{81}{16}\]
johnweldon1993
  • johnweldon1993
Make sense? the logic not the process...the process is new of course you wouldnt get that perfectly yet
rebeccaxhawaii
  • rebeccaxhawaii
BUT WHY
johnweldon1993
  • 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
  • rebeccaxhawaii
can you change the rules please
rebeccaxhawaii
  • rebeccaxhawaii
but kay i get it at its best extent of my understanding
rebeccaxhawaii
  • rebeccaxhawaii
math cheating
johnweldon1993
  • 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
  • rebeccaxhawaii
end ... of time until the world blows up
rebeccaxhawaii
  • rebeccaxhawaii
okay i got it wait why is it ^2
johnweldon1993
  • 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
  • rebeccaxhawaii
WHAT
johnweldon1993
  • 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
  • rebeccaxhawaii
okay i got it
johnweldon1993
  • johnweldon1993
You sure? Okay... So we have \[\large \frac{y}{2} + \frac{49}{16} = (x + \frac{9}{4})^2\]
johnweldon1993
  • 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
  • johnweldon1993
*Rid of..not right of*
rebeccaxhawaii
  • rebeccaxhawaii
is that it
johnweldon1993
  • johnweldon1993
mmhmm that's it :)
rebeccaxhawaii
  • rebeccaxhawaii
yay
rebeccaxhawaii
  • 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
  • rebeccaxhawaii
last 2 questions
johnweldon1993
  • 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
  • rebeccaxhawaii
5x^2+9x+3
johnweldon1993
  • 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
  • rebeccaxhawaii
woah
johnweldon1993
  • johnweldon1993
Mmhmm, buts its easy, just plug that into a calculator and you're good :D
rebeccaxhawaii
  • rebeccaxhawaii
okay so those are the soluutions why are there 2
johnweldon1993
  • 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
  • rebeccaxhawaii
bare with me slowly
johnweldon1993
  • johnweldon1993
I'm here love :D
rebeccaxhawaii
  • rebeccaxhawaii
wdym now just solve that
johnweldon1993
  • 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
  • rebeccaxhawaii
okay LAST ONE HOW EXCITING
rebeccaxhawaii
  • rebeccaxhawaii
HELLO
johnweldon1993
  • 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
  • johnweldon1993
Sorry a lot to type! lol
rebeccaxhawaii
  • rebeccaxhawaii
omg no you are literally a saint taking the time to educate the uneducated gosh you are an angel thank you
johnweldon1993
  • johnweldon1993
Lol I got you :D
johnweldon1993
  • johnweldon1993
Does everything make sense *for the most part anyways* Is Professor McMerlock happy? ;D
rebeccaxhawaii
  • rebeccaxhawaii
haha yes i am thank you so much until next time

Looking for something else?

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