Question

Please help :( How would i do this??


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.




Just the steps, or how i would do this, please. :) Please - Thank you! ♥

Answer 1

They have to be in standard form too. But, it is ok if they are not. I can do that myself. :)

Answer 2

@♥Love.Missy24♥ Have you studied quadratic equations?

Answer 3

Yea - but i don't recall all of it. Plus i have never been good at math. I only passed by like a few points.. :/

Answer 4

no problem, you can always learn it.
A quadratic equation is of the form
\[y=ax^2+bx+c\]
Solution of quadratic equation is the values of x for which y=0
and this is also a function, we can replace y by f(x), just a denotion
\[f(x)=ax^2+bx+c\]
Do you follow till here?

Answer 5

Yea- i think i got that part! ^.^

Answer 6

Cool, now solution of a quadratic equation is given by this formula
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
this will give us two values, on using + and other using -

Answer 7

*one

Answer 8

Okay - i think i remember that part - i didn't like it lol :) So, how would this work with making the function with two real rational solutions?

Answer 9

yes, let's make our equation easy
I'll put a=1
b=0
\[f(x)=x^2+c\]
Let's put f(x) =0, to find value of x in terms of c, can you try?

Answer 10

Ok so it would all be equal to 0

Answer 11

0 = 1x^2 + 0x + c ?

Answer 12

yes, can you find x in terms of c from this
\[x^2+c=0\]

Answer 13

? I do not think so. What would be the first step in solving that?

Answer 14

Sorry :/

Answer 15

subtract c from both sides

Answer 16

so it would be x^2 = 0 - c

Answer 17

yes
\[x^2=-c\]
Now take root both sides

Answer 18

ok so x = c

Answer 19

oh my bad c^2

Answer 20

you have to take square if
\[x^2=4\]
then\[x=\pm \sqrt 4\]
do the same thing here

Answer 21

oh? so it would be (i do not know how to make it show like that) x = +-sqrt(c) ??

Answer 22

yes, it's ok
But would you have c in the square root or something else?

Answer 23

In the square root. Am i on the right track - or so far out in left field :/ Sorry if i made this complicated.

Answer 24

No problem, I mean we had x^2=4
so I wrote x=+- sqrt (4)
now we have
\[x^2=\underline {- c}\]
There should be something different, shouldn't it?

Answer 25

yea.

Answer 26

So what would be the value of x

Answer 27

4 ?

Answer 28

How do you say 4?

Answer 29

cause you said x^2 = 4

Answer 30

the way to do it is by considering the polynomial:

\[f(x) = ax^2 + bx + c = 0\]
and knowing that the x values/solution are found with the quadratic formula:\[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]

we are only going to observe the term:\[\pm \sqrt{b^2 - 4ac}\]

two real rational solutions: when b^2 - 4ac > 0 and sqrt(b^2 - 4ac) is rational
two real irrational solutions: when b^2 - 4ac > 0 and sqrt(b^2 - 4ac) is irrational
two complex solutions: when b^2 - 4ac < 0 [leading to a negative under the root]

lmk if you have questions. i'm pretty sure this is how they want you to do it.

you'll just need to find a,b and c values which satisfy the conditions

Answer 31

Even if x^2=4. x would be the root of 4.
That was just for explanation
\[x^2=-c=> x=?\]

Answer 32

Thank you guys very much. I got it - kinda - i will ask you guys if i have any more questions. ♥ Thank you so much!

Answer 33

Did you understand?

Answer 34

Yea - kinda - but i have my parents yelling at me about this and i just cant think right now :/ I will ask if i have any more questions.

Answer 35

Aww:( yes, it would take a bit of work to understand this. Come here back when you have a free mind