Mathematics
OpenStudy (anonymous):

Put the 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.

OpenStudy (anonymous):

f(x) = (x-3)(x-2) g(x) = (x - sqrt(2)(x-sqrt(3)) h(x) = (x-2i)(x-3i) Do those match that? If not, could you give me some brand new functions? :/

ganeshie8 (ganeshie8):

first two are okay, u can simplify them and put in standard form

ganeshie8 (ganeshie8):

third function is NOT OK. here is the hint :- complex roots come in conjugate pairs

OpenStudy (anonymous):

But the last looks correct? Wait, so would I switch it to (x+3i)

OpenStudy (anonymous):

Or would it be (x+2i)? @ganeshie8

ganeshie8 (ganeshie8):

yup! h(x) = (x-2i)(x+2i)

OpenStudy (anonymous):

Cool! Could you help me with three more regarded the same thing? (I have a set of 4 questions)

ganeshie8 (ganeshie8):

shoot

OpenStudy (anonymous):

(sorry, this is still part of #1 and I'm clueless when it comes to functions) Create these three functions and explain to how you know these functions meet each condition. (Hint: Make sure that the b is even on g(x).)

OpenStudy (anonymous):

I feel, though, you've already given me this answer, but in a simpler form XD also, for 1, 2, & 3, are they in standard form? s: because I need that..

ganeshie8 (ganeshie8):

nope, they're not in standard form. u need to multiply and simplify them

ganeshie8 (ganeshie8):

standard form :- \(\large ax^2+bx+c\)

OpenStudy (anonymous):

1. f(x)=x^2 - 5x+6 2. i'm lost on this because of the square root... is it g(x)= -sqrt(2)x+x+sqrt(6)? 3. h(x) = x^2+4

OpenStudy (anonymous):

@ganeshie8

ganeshie8 (ganeshie8):

knw what, our 2nd equation is also incorrect

ganeshie8 (ganeshie8):

irrational roots also come in pairs !

ganeshie8 (ganeshie8):

g(x) = (x - sqrt(2)) (x+sqrt(2)) now try :)

OpenStudy (anonymous):

g(x)=x^2-2 ?

OpenStudy (anonymous):

hello....

hartnn (hartnn):

yes, that g(x) is correct....

hartnn (hartnn):

Create these three functions and explain to how you know these functions meet each condition. (Hint: Make sure that the b is even on g(x).)" so, now actually solve those quadratic equations , and see whether you get 1) the roots as real rational for x^2-5x+6 similarly for 2nd and 3rd

hartnn (hartnn):

f(x) = x^2-5x+6 = (x-3)(x-2) = 0 so, x-3 = 0, x-2 =0 x=3,2 <----real rational hence verified

OpenStudy (anonymous):

Okay! I think I got this step on my own :3 I'll just show my work on how to solve it! #2: 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.

hartnn (hartnn):

vertex form = a(x+h)^2 + k

hartnn (hartnn):

a=1, because we have 1x^2

OpenStudy (anonymous):

1. x^2 + 6 = f(x) + 5x (i dont know how to make it look like a(x+h)^2+k 2. (i dont know for this) 3. same with #2

OpenStudy (anonymous):

I'm still here but I'm going to continue with my answer for #1 on my worksheet

hartnn (hartnn):

the Question only asks for f(x) , no need to convert all 3 so, f(x) = x^2-5x+6 do you know how to complete the square ?

OpenStudy (anonymous):

o.o oh... yeah.

OpenStudy (anonymous):

Bleh, I know how but the equation is wonky. It doesn't compare to what's given in the lesson

hartnn (hartnn):

ok, yes our f(x) is not that easy now we can continue with this f(x) OR we can select new f(x) from the beginning what u wannna do ?

OpenStudy (anonymous):

a leading coefficient of 1 must exist on the x2 term

hartnn (hartnn):

here it does

OpenStudy (anonymous):

Divide the coefficient on x and square the result ?

hartnn (hartnn):

yeah, here its 5 divide by 2, so 5/2 square it what u get ?

OpenStudy (anonymous):

OH. We're going from standard, I forgot. I was looking at the... function before standard. 25/4 = 6.25

hartnn (hartnn):

keep it as 25/4 so add and subtract 25/4 f(x) = (x^2-5x +25/4) +(6-25/4) got this ?

OpenStudy (anonymous):

so far...

hartnn (hartnn):

whats 6-25/4 = .. ?

hartnn (hartnn):

(x^2-5x +25/4) = (x-5/2)^2 ----->(x+h)^2 form

OpenStudy (anonymous):

sorry s: = 1/4

OpenStudy (anonymous):

= -1/4**

hartnn (hartnn):

yes, so f(x) = (x-5/2)^2 -1/4 which is in the form a(x+h)^2 +k a=1, h=-5/2, k=-1/4, done! :)

OpenStudy (anonymous):

f(x)=(x-5/2)^2+1/4? --- oops XD is that it? that's all there is to it? :O well, okay. I'd give you the medal if i hadn't given the other person it. #4 (last one!!) 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.

OpenStudy (anonymous):

i know we just did completing the square but the discriminant part throws me off.. what is that?

hartnn (hartnn):

wait, f(x) = (x-5/2)^2-1/4

OpenStudy (anonymous):

but the equation is +k ?

OpenStudy (anonymous):

lol, i skipped a question. I thought we did #3

hartnn (hartnn):

yeah, but here k is negative, k = -1/4 and it can be negative...... the discriminant \(b^2-4ac\) will be negative only if the roots are complex , so for this we must take h(x) only

OpenStudy (anonymous):

o.o you lost me.. (i get that it can be negative, okay. i just thought it would be somehow converted to positive)

hartnn (hartnn):

f(x) = (x-5/2)^2-1/4 is correct, keep it this way

OpenStudy (anonymous):

gotcha. thats what i have

OpenStudy (anonymous):

solving solutions of g(x) for g(x)=x^2-2

OpenStudy (anonymous):

#4 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.

OpenStudy (anonymous):

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