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 February 9th, 2013, 08:38 AM #1 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 What is the correct way to apply quadratic formula? Hello, I have this equation which I want to solve for variable $t$: $y= v*t*sin(h)-\frac{1}{2}g*t^2$ This is the Quadratic equation which I was taught in school: $x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$ So I substitute the a, b and c: $t = \frac{ -v*sin(h)\pm sqrt{v^2*sin^2(h)-2*g*y} }{g}$ The problem is that Wolfram Alpha says the above is incorrect, and this is correct: $t = \frac{ v*sin(h)\pm sqrt{v^2*sin^2(h)-2*g*y} }{g}$ Click here to see the solution at Wolfram Alpha I would like to learn why Wolfram Alpha is saying this. What do you think? Thank you for your time. Kind regards, Marius
 February 9th, 2013, 09:13 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: What is the correct way to apply quadratic formula? You need to put your equation in standard form first: $$$\frac{1}{2}g$$t^2+$$-v\sin(h)$$t+(y)=0$ Now, you can see: $a=\frac{1}{2}g,\,b=-v\sin(h),\,c=y$
 February 9th, 2013, 09:59 AM #3 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 Re: What is the correct way to apply quadratic formula? Thanks.M
February 9th, 2013, 06:09 PM   #4
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From: Ottawa Ontario, Canada

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Re: What is the correct way to apply quadratic formula?

Quote:
 Originally Posted by king.oslo $y= v*t*sin(h)-\frac{1}{2}g*t^2$
Make it easier on yourself, King; let sin(h) = k, and multiply through by 2:
2y = 2vtk - gt^2
Then:
gt^2 - 2vkt + 2y = 0
Get my drift?

February 12th, 2013, 05:28 AM   #5
Senior Member

Joined: Sep 2010
From: Oslo, Norway

Posts: 162
Thanks: 2

Re: What is the correct way to apply quadratic formula?

Quote:
Originally Posted by Denis
Quote:
 Originally Posted by king.oslo $y= v*t*sin(h)-\frac{1}{2}g*t^2$
Make it easier on yourself, King; let sin(h) = k, and multiply through by 2:
2y = 2vtk - gt^2
Then:
gt^2 - 2vkt + 2y = 0
Get my drift?
Yes Denis, that makes sense.

I was looking at the dimensions in the result, and I cannot get them to match.

$y= v*t*sin(h)-\frac{1}{2}g*t^2$

$m= \frac{m}{s}*s-\frac{m}{s^2}*s^2$
$m= m-m$
This looks right

What about this?
$t = \frac{
v*sin(h)\pm sqrt{v^2*sin^2(h)-2*g*y}
}{g}$

$s = \frac{
\frac{m}{s}*sqrt{(\frac{m}{s})^2-\frac{m}{s^2}*m}
}{\frac{m}{s^2}$

$s = \frac{
\frac{m}{s}*sqrt{(\frac{m}{s})^2-(\frac{m}{s})^2}
}{\frac{m}{s^2}$

$s = \frac{
\frac{m}{s}*sqrt{(\frac{m}{s})^2}
}{\frac{m}{s^2}$

$\text{ Above I omitted the one of the two (\frac{m}{s})^2 inside the quare root because they have the same unit, that way I can take the quare root next:}$

$s = \frac{
\frac{m}{s}*(\frac{m}{s})
}{\frac{m}{s^2}$

$s = \frac{
\frac{m^2}{s^2}
}{\frac{m}{s^2}$

Everything but the one m cancel out, and I am left with:



Where did I go wrong?

Thanks M

 February 12th, 2013, 06:21 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,937 Thanks: 2210 You incorrectly replaced "±" with "*" early on.
 February 12th, 2013, 06:34 AM #7 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 Re: What is the correct way to apply quadratic formula? Skipjack, thank you so much! Now everything works out in my lovely mathematics Thank you!M
 February 13th, 2013, 04:48 AM #8 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 Re: What is the correct way to apply quadratic formula? Another question: When I am picking $a$, $b$ and $c$, I have the choice to pick the inverted set of numbers, because the right hand side of the standard form is 0 anyway. Are there any implications of doing this? I think so, at least it seems that for one set of numbers I may end up with a complex number as the result, but not in the other. For example, I suspect $x$ in the case of: $a = n_1 b = -n_2 c = -n_3$ may not always be equal to $x$ in the case of: $a = -n_1 b = n_2 c = n_3$ Is this the case? If so, what is the correct way to pick the sign of $a$, $b$ and $c$? Thank you for your time. Kind regards, Marius
 February 13th, 2013, 05:52 AM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: What is the correct way to apply quadratic formula? $\frac{n_2\,\pm\,\sqrt{(-n_2)^2\,-\,4(n_1)(-n_3)}}{2n_1}\,=\,\frac{-n_2\,\mp\,\sqrt{n_2^2\,-\,4(-n_1)(n_3)}}{-2n_1}$
February 13th, 2013, 06:02 AM   #10
Senior Member

Joined: Sep 2010
From: Oslo, Norway

Posts: 162
Thanks: 2

Re: What is the correct way to apply quadratic formula?

Quote:
 Originally Posted by greg1313 $\frac{n_2\,\pm\,\sqrt{(-n_2)^2\,-\,4(n_1)(-n_3)}}{2n_1}\,=\,\frac{-n_2\,\mp\,\sqrt{n_2^2\,-\,4(-n_1)(n_3)}}{-2n_1}$
Okay, thanks.M

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