Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

Calculate the sum and the product of the equation roots. Give the answer in fractions (ex: x = 3 ¾).

OpenStudy (anonymous):

\[-3x ^{2}+2x+4=0\]

OpenStudy (anonymous):

do you want to find the answer when you plug x in the equation?

OpenStudy (anonymous):

or just factor?

OpenStudy (anonymous):

that is what I dont know, because this question came as it is, I know how to do quadratic equations, but I just dont understand what the question wants from me

OpenStudy (anonymous):

It wants you to understand the relationship between the roots and coefficients in a quadratic.

OpenStudy (anonymous):

If you do, you can read off from that equation what you need to know without any working (well, much working).

OpenStudy (anonymous):

how so?

OpenStudy (anonymous):

yeah, how so?

OpenStudy (anonymous):

If you have the quadratic equation,\[x^2+bx+c=0\]if\[\alpha \]and\[\beta \]are the roots, then you'd also have\[(x-\alpha)(x-\beta)=0\]If you expand this, you get\[x^2-(\alpha + \beta)+{\alpha}{\beta}=0\]Comparing the first and second equations, you have that they will be identical equations if and only if their coefficients are identical. Then\[x^2+bx+c=x^2-(\alpha + \beta) + {\alpha}{\beta}\]if and only if\[b=-(\alpha + \beta)\]and\[c={\alpha}{\beta}\]

OpenStudy (anonymous):

The coefficient of x is related to the sum of roots, and the coefficient of x^0 (i.e. c) is related to the product of the roots.

OpenStudy (anonymous):

Your equation differs only in that it has a coefficient for x^2 (i.e.-3). But guess what? You can get rid of it by dividing your equation through by -3.

OpenStudy (anonymous):

ok.... *eyes wide open, helpless face*

OpenStudy (anonymous):

LOL, what he's trying to say is you can get the equation you have in the form of the general equation when you divide by -3 :)

OpenStudy (anonymous):

\[-3x^2+2x+4=0 \iff x^2-\frac{2}{3}x-\frac{4}{3}=0\]

OpenStudy (anonymous):

Then you can read off the sum and product of roots.

OpenStudy (anonymous):

\[\alpha + \beta = \frac{2}{3}\]\[{\alpha}{\beta}=-\frac{4}{3}\]

OpenStudy (anonymous):

I think he fainted ._.

OpenStudy (anonymous):

lol you ok bmf?

OpenStudy (anonymous):

loki what were you trying to say, can you summarize it please? :)

OpenStudy (anonymous):

more like in 5 steps? please?

OpenStudy (anonymous):

ok, just recovered

OpenStudy (anonymous):

lol, he'll summarize it don't worry ^_^

OpenStudy (anonymous):

i am starting to understand the idea

OpenStudy (anonymous):

great

OpenStudy (anonymous):

Imagine alpha and beta are the roots of some quadratic equation. Then you can write the quadratic as a product of its factors\[(x-\alpha)(x-\beta)\]Expand and see that it equals\[x^2-(\alpha + \beta)x + {\alpha}{\beta}\]So, you should see from the form that the coefficient of x in a quadratic is related to the SUM of the roots, while the constant is related to the PRODUCT of the roots.

OpenStudy (anonymous):

so now he can put them equal to each other and continue.

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

okay, problema solved ^_^

OpenStudy (anonymous):

the only catch is that you have to ensure the coefficient of x^2 is 1.

OpenStudy (anonymous):

he'll divide by -3 like you've said , got that andy?

OpenStudy (anonymous):

but that's easily handled - factor it out.

OpenStudy (anonymous):

ok, got it, nut I think I have to get drunk like loki so next time I can figure stuff faster

OpenStudy (anonymous):

*but not nut

OpenStudy (anonymous):

When you're just given the quadratic\[ax^2+bx+c\]and it's not set equal to anything, to get at the relationship for the roots, you have to factor out, rather than divide (there's nothing on the other side of an equals sign to divide). So in this case,\[ax^2+bx+c=a[x^2+\frac{b}{a}x+\frac{c}{a}]\]and you'd relate the roots to the stuff in the brackets.

OpenStudy (anonymous):

lol, thank you loki ^_^

OpenStudy (anonymous):

yeah, have a beer and mathematics will be more fun ;)

OpenStudy (anonymous):

you don't need to get drunk to answer a question andrius

OpenStudy (anonymous):

lol so he'll get 4 when you ask him 1+1 ._.

OpenStudy (anonymous):

with any luck...

OpenStudy (anonymous):

well, I can see loki is smart, and he just said I should have a beer and maths will be more fun, so, I guess I have nothing to lose

OpenStudy (anonymous):

i trust loki

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

just your brain cells ._. LOL

OpenStudy (anonymous):

drinking is bad bad bad, but you boys are just so stubborn to accept that.

OpenStudy (anonymous):

aw...

OpenStudy (anonymous):

I'm serious, men always do the opposite lol, wanting to try everything

OpenStudy (anonymous):

well women too, but not as much though

OpenStudy (anonymous):

depends on the girl actually ^^"

OpenStudy (anonymous):

still though.

OpenStudy (anonymous):

what happens if you leave raw chicken out of the fridge for two days but the weather's cold? do you think it's safe to cook?

OpenStudy (anonymous):

it depends, you must check if there's any fungus left on top , smell it ; if it's good then yes, if not then you must imediately throw it away

OpenStudy (anonymous):

cool

OpenStudy (anonymous):

wait, if the weather's cold, then I guess it's safe

OpenStudy (anonymous):

it'll keep the chicken cooled just like leaving it in a fridge.

OpenStudy (anonymous):

there's no difference, the only small difference is temperature and where you place it.

OpenStudy (anonymous):

why ask such a question anyway?

OpenStudy (anonymous):

I am kind of lost on how beer, chikens and men wanting to do the opposite relate to each other

OpenStudy (anonymous):

exactly lol

OpenStudy (anonymous):

see what happens when you drink? you lose the relation between subjects and things.

OpenStudy (anonymous):

lol...sad

OpenStudy (anonymous):

evidence = loki's question , no more drinking for you

OpenStudy (anonymous):

but he has a point, what does that question got to do with drinking?

OpenStudy (anonymous):

ok, well, to be truthful...

OpenStudy (anonymous):

I never drink

OpenStudy (anonymous):

ever

OpenStudy (anonymous):

it doesn't. it's just that i spent $6 on some chicken but then went out for dinner and i'm wondering if i should now cook it before bed or throw it out?

OpenStudy (anonymous):

lol, good boy

OpenStudy (anonymous):

does it look fresh? does it smell funny?

OpenStudy (anonymous):

you are a good girl too

OpenStudy (anonymous):

oh you're all good

OpenStudy (anonymous):

^_^

OpenStudy (anonymous):

lol, so are you

OpenStudy (anonymous):

i think i'll cook it then go to bed

OpenStudy (anonymous):

you can cook?

OpenStudy (anonymous):

sorry to hijack the thread bmf

OpenStudy (anonymous):

not really. i push buttons on the microwave

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

atleast that makes us all equal?

OpenStudy (anonymous):

and I ask my neighbour to cook food for me

OpenStudy (anonymous):

atleast I can cook an egg

OpenStudy (anonymous):

better than nothing lol ^_^

OpenStudy (anonymous):

though I'm still learning, I want to >_<

OpenStudy (anonymous):

no, I am kidding, I can actually make any kind of food, at one time I even considered becoming a cook LOL. so, now I can bake cakes, complex meals and all kind of desserts

OpenStudy (anonymous):

oh lucky you, I just burn everything lol

OpenStudy (anonymous):

still learning :)

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

anyways, basket time, talk later

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

bye

OpenStudy (anonymous):

i;m out...bye

OpenStudy (anonymous):

Try www.aceyourcollegeclasses.com

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!