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Topic review (newest first)

2012-12-13 18:02:27


Glad you worked it out yourself.

2012-12-13 17:02:02

Hi Bob:
That website shows exactly how I have done it but, obviously I have made some mistake if I did not get my 8.

Hi Bobbym:
Yes, I know it couldn't be -3 (because I am trying to find seconds here, -3 will not be it) and definately would not be 1 second. I'll show you my work using the website Bob has given.


-4.9t^2 + 24.5t + 117.6 = 0

-24.5 (sqr)24.5^2 - 4(-4.9)(117.6) / 2(-4.9)
First I did all of the multiplication with parenthesis:

-24.5 (sqr)24.5^2 - 4(-576.24) / -9.8

Then I multiplied the number in the parentheses with 4 and multiplied 24.5^2 by itself:

-24.5 (sqr)600.25 - (-2304.96) / -9.8

Since the number in the parenthesis is a negative and there is a minus, I changed it to a plus:

-24.5 (sqr)600.25 + 2304.96 / -9.8

-24.5 (sqr)2905.21 / -9.8

Now I square the number and create two equations:

-24.5 + 53.9 / -9.8
-24.5 - 53.9 / -9.8

29.4 / -9.8 = -3
-78.4 / -9.8 = 8

It seems as if I got it right this time! It must of been a small mistake I carelessly made along the way. I think I might have divided first, the add/subtract. Thank you for your time and help guys!wink

2012-12-13 07:57:14

Actually it is the last one:
-4.9t^2 + 24.5t + 117.6 = 0

Sorry, that is the one I meant. The equation is right but the roots are obviously not. Like I said the ball can not possibly hit the ground in one second so you know something is wrong. The quadratic formula can be a little tricky.

Always plot first, you can see the roots now.

bob bundy
2012-12-13 03:53:54

hi demha,

That equation looks right to me but your answers of -3.30 and 1.22 don't work in it.

Are you using the quadratic formula?

You'll need to write your steps if you want someone to find the error.

Have a look at … olver.html


2012-12-13 01:29:07

Actually it is the last one:
-4.9t^2 + 24.5t + 117.6 = 0

First time I did it, I got a postive 3 in the end and chose that as my answer. I must have made a mistake somewhere then?

2012-12-12 21:47:32


Did you pick the third equation

as you said?

You know that neither of those answers make any sense. How could the ball hit the ground in 1 second? So either you picked the wrong equation or you did not solve it correctly.

Answers in math ought to make sense. Mine don't but that is only because I work on nonsense.

2012-12-12 21:35:41

My class shows me how to do #15, this website shows the exact same thing: since I can't post links, search on Google: Using Quadratic Formula to Find the Zeros of a Polynomial  and click the first link.

Here are my answers:
-3.30 (round up becomes -3)
1.22 (round up becomes 1)

bob bundy
2012-12-12 19:03:45

hi demha,

That is correct.  smile


2012-12-12 18:23:12

Alright, let me try that my self and please do tell any mistakes I make along the way. I will also try #15.

x = 3 and x = -7

(x – 3) (x + 7)
x(x – 3)
7(x - 3)

(x^2 – 3x)
(7x - 21)

When adding together, you are taking away 7x from 3x which becomes -4x. This changes the sign and that means the correct equation is:

x^2 + 4x – 21

I see where I made my mistake, I didn’t change the sign!

2012-12-11 23:19:34

Your teacher is not correct but if he/she wants to say you are right then let's leave it alone for now.


Work backwards to write a quadratic equation that will have solutions of x = 3 and x = -7.

The correct equation is,

2012-12-11 23:09:21

anonimnystefy wrote:

Hi bobbym

Her 4) and 13) are also correct.

Hi anonimnystefy,
I'm a guy... so it would be a he, not a she lol


Hi bobyym,
Thanks for answering. I submited my work already though and it seems that all but #6 and #15 are correct (according to my teacher). Right now she wants me to show her the work I have done to get those answers. I believe those two may be wrong.

2012-12-11 20:30:05

Can't be one positive and one negative.

2012-12-11 20:25:57

13) says integers...

2012-12-11 20:21:34


Four is missing a root? 13 asks for positive numbers.

2012-12-11 20:18:44

Hi bobbym

Her 4) and 13) are also correct.

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