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Mathematics 21 Online
zarkam21 (zarkam21):

Help please

zarkam21 (zarkam21):

OpenStudy (skullpatrol):

Any ideas?

OpenStudy (mathmale):

Please just post your question right off. Not necessary to write, "help please." do whatever you have to do to move both terms from the right side of the equation to the left side. Your result?

zarkam21 (zarkam21):

actually @mathmale I did ..but you never responded so I decided to close that question and repost

OpenStudy (skullpatrol):

What is the definition of "standard form" for a polynomial?

OpenStudy (skullpatrol):

Do you know how to arrange the terms in decreasing degree of the variable?

zarkam21 (zarkam21):

Is the answer? \[2x^2-3=-2(3x+1)\]

zarkam21 (zarkam21):

@skullpatrol

jimthompson5910 (jim_thompson5910):

Similar example: Let's say we had \[\Large 5x^2 + 10 = 4x - 7\] and we wanted to get that into standard form. The goal is to get everything to one side. Then combine like terms. In this example, that means we subtract 4x from both sides first \[\Large 5x^2 + 10 = 4x - 7\] \[\Large 5x^2 + 10-4x = 4x - 7-4x\] \[\Large 5x^2 + 10-4x = 4x-4x - 7\] \[\Large 5x^2 + 10-4x = 0x - 7\] \[\Large 5x^2 + 10-4x = 0 - 7\] \[\Large 5x^2-4x + 10 = -7\] Then we add 7 to both sides \[\Large 5x^2-4x + 10 = -7\] \[\Large 5x^2-4x + 10+7 = -7+7\] \[\Large 5x^2-4x + 17 = 0\] and that wraps up the problem. Keep in mind that this is just an example.

zarkam21 (zarkam21):

Oka so would I be correcT?

zarkam21 (zarkam21):

@3mar

jimthompson5910 (jim_thompson5910):

that's not in standard form @zarkam21

jimthompson5910 (jim_thompson5910):

standard form is ax^2 + bx + c = 0 where a,b,c are place holders for numbers Something like 3x^2 + 12x - 19 = 0 is standard form. It's not the answer, but the answer will look like this

OpenStudy (skullpatrol):

One side of the equation is 0 and the other is arranged in decreasing degree of the variable.

zarkam21 (zarkam21):

And then

OpenStudy (skullpatrol):

And then that is your answer :-)

OpenStudy (mathmale):

If you want to go through this faster, please move the 2 terms from the right side to the left side. Show your result. Should be easy since you've done this before.

zarkam21 (zarkam21):

Okay one sec I'm just going to do this all out on paper then I will let you know what I get =)

zarkam21 (zarkam21):

@skullpatrol =)

OpenStudy (mathmale):

Prefer you show one step at a time.

OpenStudy (skullpatrol):

Paper is good =)

OpenStudy (mathmale):

I'll let skullpatrol take over here.

OpenStudy (skullpatrol):

Nah @mathmale you can help please :-)

zarkam21 (zarkam21):

I would add 3 to both sides first right?

OpenStudy (skullpatrol):

Yes

OpenStudy (mathmale):

Well, that approach would work, but you've chosen not to do what I first suggested. Leave the left two terms where they are and move the right two terms over to the left side. How would you do that?

zarkam21 (zarkam21):

after that I would get \[2x^2=-6x\]

zarkam21 (zarkam21):

\[2x^2=-6x+1\]

OpenStudy (skullpatrol):

good

OpenStudy (skullpatrol):

Now what?

OpenStudy (mathmale):

You're not at liberty to discard terms. You have 2x^2-3=-6x-2. Add 6x to both sides. then ADD 2 to both sides. You will end up with 4 terms on the left. 2 of them can be combined. Mind doing that?

zarkam21 (zarkam21):

\[2x^2+6x-3=1 ?\]

OpenStudy (mathmale):

Your \[2x^2=-6x+1\] will "work," but convention says we usually consolidate terms on the left, not on the right. Again, i am going to excuse myself, rather than confuse you with 2 different approches to solving the same problem. Good luck.

zarkam21 (zarkam21):

Okay =)

zarkam21 (zarkam21):

@skullpatrol

OpenStudy (skullpatrol):

Any ideas?

zarkam21 (zarkam21):

now would I add 6x to both sides?

OpenStudy (skullpatrol):

Do it please :-)

OpenStudy (mathmale):

Yes. But I gave you clear and specific instructions for the solution of this problem, long ago.

OpenStudy (skullpatrol):

Learn by doing ;-)

zarkam21 (zarkam21):

2x^2+6x+2-3=1

OpenStudy (skullpatrol):

All you are required to do is satisfy the definition of the "standard form" of a polynomial.

OpenStudy (mathmale):

And you are getting closer to that goal. Your standard form will look like y=ax^2+bx+c.

OpenStudy (skullpatrol):

^ y = 0

zarkam21 (zarkam21):

so next would i simply do 2-3?

OpenStudy (skullpatrol):

Yes

OpenStudy (mathmale):

:)

OpenStudy (skullpatrol):

Please take over @mathmale my battery is getting low

zarkam21 (zarkam21):

2x^2+6x-1=1

OpenStudy (mathmale):

Yes, but is that in std. form? you don't want anything on the right side. Actually, zarkam, you want an equation with zero (0) on the right side.

OpenStudy (mathmale):

You could accomplish this by subtracting 1 from both sides of your equation. Here you go: 2x^2+6x-1-1=1-1. Please simplify this.

zarkam21 (zarkam21):

Okay so what would be the next step, I'm not quite sure

OpenStudy (mathmale):

I just gave it to you. See above.

zarkam21 (zarkam21):

2x^2+6x-2=0?

OpenStudy (mathmale):

Yes. That was what you were supposed to do. This equation is in standard form. Any questions about this procedure?

zarkam21 (zarkam21):

Nope thanks =)

OpenStudy (mathmale):

Good luck. Hope you'll be patient with me next time. Take care.

zarkam21 (zarkam21):

Sure will and I apologize also, See ya soon

OpenStudy (skullpatrol):

Also study @Jim_Thompson5910's example please :-)

zarkam21 (zarkam21):

Will do =)

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