Question

Rewrite y = x2 + 14x + 29 in general form. (1 point)
Rewrite y = 3x2 - 24x + 10 in general form. (1 point)
Solve for x: (x - 9)2 = 1 (1 point)
Solve for x: x2 + 24x + 90 = 0 (1 point)
Solve for x: 2x2 - 4x - 14 = 0 (1 point)

Answer 1

http://www.purplemath.com/modules/parabola.htm
look torward the bottom

Answer 2

I need the explanation please.

Answer 3

Guys we help on this site, we do not give answers nor do we do your homework for you.

Answer 4

http://openstudy.com/code-of-conduct

Answer 5

Ok well you mind your own business please because as you can see I asked for a explanation you illiterate bum.

Answer 6

I was referring to the fact that you posted more than one question, besides you didn't do anything against the CoC @elizabethgamez did.

Answer 7

Was is that I don't understand what you mean.

Answer 8

Yes, but you should post one questions per post but other than that you didn't do anything wrong :)

Answer 9

Oh I apologize for my accusation dear sir.

Answer 10

It's fine I don't mind :)

Answer 11

Thank you.

Answer 12

\[y = ax ^{2} + bx + c\]\[y = a \left( x ^{2} + \frac{ bx }{ a } \right) + c\]\[y = a \left( x ^{2} + \frac{ bx }{ a } + \frac{ b ^{2} }{ 4a ^{2} } \right) + c - \frac{ b ^{2} }{ 4a }\]\[y = a \left( x + \frac{ b }{ 2a } \right)^{2} + c - \frac{ b ^{2} }{ 4a }\]

Answer 13

Ok can you help me solve one or two of them or at least check two and solve one

Answer 14

ok. To start, you identify your "a", "b", and "c" and substitute into the last line of the formula. Try one and I'll check it.

Answer 15

1. is y= 3x^2 - 24x + 10
2. Is y equal 3(x-4)^2-38 and solve number 5 please

Answer 16

I mean explain it to me. :)

Answer 17

ok. Let's take a look at:

y= 3x^2 - 24x + 10

Here, this is in "standard form" of:

y = ax^2 + bx + c

So, a = 3, b = -24, and c = 10 So, substituting into my final line:

\[y = 3\left( x - \frac{ 24 }{ (2)(3) } \right)^{2} + 10 - \frac{ (-24)^{2} }{ (4)(3) }\]and now just simplify.

Answer 18

but was I write or no

Answer 19

So but bear with me

Answer 20

Yes, you got that right. Good job!

Answer 21

and the second one

Answer 22

is it standard form or general form

Answer 23

Question #5 isn't about general or standard form. It's just about solving for what "x" values will give "0" on the right side. It's a standard form where "y" is already set to "0".

2x^2 - 4x - 14 = 0 -> 2(x^2 - 2x - 7) = 0

and now use the quadratic formula:\[x = \frac{ -b \pm \sqrt{b ^{2} - 4ac} }{ 2a }\]

Answer 24

ohh but 1 and 2 is general form correct

Answer 25

The form in the problem statement for #1 and #2 are in standard form, but they are asking for those to be re-written in general form, yes.

As for #5, you just plug those a, b, and c into the quadratic formula and simplify.

a = 1, b = -2, c = -7

Answer 26

So the answers for 1 and 2 are correct yes or no ;) and that will be all. Thank you again for your concern.

Answer 27

#2 you and I already did and I said that was right. As for #1:

y = (x + 7)^2 - 20

Answer 28

ok I appreciate it soooooooooooooooooooooooooooooooooooooooo much Thank youuuuuuuuuuuu I am great full. I hope you have a nice day and again thank you. Bye

Answer 29

u2! Really nice working with you! @magbak