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4,9t^2+39t+125=0
- Thread starter sunrise
- Start date
D
Deleted member 4993
Guest
Hi
I have an equation problem and i dont know way of solving it.
I had a physic problem i solved then when problem came to hear
i couldnt solve of course
4,9t^2+39t+125=0
may you help me?
Thank you
This is a Quadratic Equation".
Look it up......
D
Deleted member 4993
Guest
Yes you are right,
quadratic equation can you show me how i will solve?
Thank you
For a quick review, got to:
http://www.purplemath.com/modules/solvquad.htm
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
Did you not realise that the "quadratic equation" in Subhotoh Khan's first post was also a link to a website about quadratic equations? Both of his posts gave links but you complain about one and thank him for the other!
Sunrise,
The solution to a quadratic equation, ax^2+bx+c=0 will have two roots given by x=(-b +/- sqrt(b^2-4*a*c) ) /(2*a). Note that here "+/-" means plus or minus. Plus gives one root and minus gives the other root.
In your case, 4,9t^2+39t+125=0, a=4.9, b=39 and c=125, thus, x=(-39+/-sqrt(39^2-4*4.9*125) )/(2*4.9)
Note that in your case b^2-4*a*c=-929 which is negative. Thus the square root of a negative number will be imaginary.
I am guessing that either your 4,9 is not 4.9,
or
you have a sign missing somewhere in the equation
or
your physics problem is such that it is ill-posed and the solution doesn't exist.
Please check your equation.
Cheers,
Sai.
The solution to a quadratic equation, ax^2+bx+c=0 will have two roots given by x=(-b +/- sqrt(b^2-4*a*c) ) /(2*a). Note that here "+/-" means plus or minus. Plus gives one root and minus gives the other root.
In your case, 4,9t^2+39t+125=0, a=4.9, b=39 and c=125, thus, x=(-39+/-sqrt(39^2-4*4.9*125) )/(2*4.9)
Note that in your case b^2-4*a*c=-929 which is negative. Thus the square root of a negative number will be imaginary.
I am guessing that either your 4,9 is not 4.9,
or
you have a sign missing somewhere in the equation
or
your physics problem is such that it is ill-posed and the solution doesn't exist.
Please check your equation.
Cheers,
Sai.
Hi
I have an equation problem and i dont know way of solving it.
I had a physic problem i solved then when problem came to hear
i couldnt solve of course
4,9t^2+39t+125=0 . . . . This is posted incorrectly. (Use the decimal point instead
of that comma that is used in certain other countries.) This equation is missing
the negative sign for the lead coefficient of the trinomial.
may you help me?
Thank you
s(t) = -4.9t^2 + 39t + 125
The distance above ground level (or the height) in meters, at time t in seconds, is s(t).
The object is launched at an initial velocity of 39 meters/second from an initial
height of 125 meters.
Your problem is meant to be to find the time t, in seconds, when the object will
reach ground level, that is, when s(t) = 0 meters.
So, you are to solve for the (positive) t value for the following equation:
-4.9t^2 + 39t + 125 = 0
Source:
http://www.purplemath.com/modules/quadprob.htm