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September 19th, 2016, 11:19 AM  #1 
Newbie Joined: Sep 2016 From: alaska Posts: 16 Thanks: 0  Hey guys. How would I go about solving this?
A falling object travels a distance given by the formula d = 2t + 16t^2, where t is measured in seconds and d is measured in feet. How long will it take for the object to travel 68 ft? I don't even know where to start lol. 
September 19th, 2016, 11:24 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
Can you solve $68 = 2t + 16t^2$ 
September 19th, 2016, 11:31 AM  #3 
Newbie Joined: Sep 2016 From: alaska Posts: 16 Thanks: 0  
September 19th, 2016, 12:04 PM  #4 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
You should not have been given this problem if you have not studied the quadratic formula. Do you understand why $68 = 2t + 16t^2$ is correct? You can rearrange that to get $16t^2 + 2t  68 = 0.$ That is STANDARD FORM for a quadratic equation. You can solve any quadratic equation in standard form using the quadratic formula. So here goes. $ t = \dfrac{\ 2 \pm \sqrt{2^2  4(16)(\ 68 )}}{2 * 16} = \dfrac{\ 2 \pm \sqrt{4 + 4352}}{32} = \dfrac{\ 2 \pm 66}{32} \implies$ $t = \dfrac{64}{32} = 2.$ 
September 19th, 2016, 12:47 PM  #5 
Senior Member Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 
to solve $16t^2+2t=68$ we simplify by 2 to get $8t^2+t34=0$ $8t^232+t2=0$ $8(t+2)(t2)+(t2)=0$ $(t2)(8(t+2)+1)=0$ $(t2)(8t+17)=0)$ thus $t=2 s$ $17/8$ can't be a solution. 
September 19th, 2016, 08:40 PM  #6  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
because there is no motivation as to why you knew to split up 34 into those particular numbers. It may be natural for you, though, if you've been doing it this way off and on for years. Also, this step inserted between the second and the third steps makes it flow better: $\displaystyle 8(t^2  4) + t  2 = 0$  

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guys, hey, solving 
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