You could write this as...0 = 9t-4.9t^2
0 = t(9-4.9t)You could write this as...
0=t ( ? )
can you fill in the "?"
I'm sorry, I can't. I do not know what you are asking...Well [imath]0=9t-4.9t^2 [/imath] is the same as [imath]4.9t^2-9t=0[/imath] or [imath]t(4.9t-9)=0[/imath].
Can you solve the last?
0 = t(9-4.9t)
But I don't see how that helps sorry...
Again, I really do not know how to solve (9-4.9t)=0...Well, if a*b = 0 then a=0 and/or b=0
So, t=0 is a solution. To find the other solution what value must t have to make (9-4.9t)=0 ?
EDIT: This idea can be extended, so if you have ( x-2 )( 2x-3 )( x-5 ) = 0, then the possible solutions are x=2, x=? and x=?
Again, I really do not know how to solve (9-4.9t)=0...
Your edit, x could also equal 1.5 and 5 right?
I get this now. To resolve,I assume that you're still struggling. How would you start to simplify the following...
( 4.9t - 9 ) = 0
4.9t - 9 = 0
Hint: now add something to both sides so that the "4.9t" will be on its own
Do you mean square rather than square root. Are you talking about the [imath] t^2 [/imath] from the original equation [imath] 0 = 9t-4.9t^2 [/imath] ?But how where does the square root go?
sorry, yes, you are quite right, the square.Do you mean square rather than square root. Are you talking about the [imath] t^2 [/imath] from the original equation [imath] 0 = 9t-4.9t^2 [/imath] ?