Radical Equation Application Problem

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u16_l3_t1_we3 Radical Equation Application Problem
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That's a neat trick with the (3/2). Usually I would do (36x3)/2 (long way I suppose).

TigerTzu
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You mention a lot of different times about making sure to check the solution when using the principal square root and how at sometimes it can be both negative and positive and other times just positive. Can you make a video about this, it is one point that has been confusing me about some of the math videos. Thanks!

jbh
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Problem seems a bit contrived. The expression given for v is a rearrangement of the more common formula for kinetic energy, E_k = (1/2)mv^2. Normally you'd just start with the more common formula, plug in values, and compute. No need to involve radicals at all.

Crissix
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Why did you multiply them by 3/2? Where did you get 3/2? Pls answer

bubbysvlog
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If anyone is interested: You don't have to 'check your answer'. You can put limitations on your solution by isolating the +sqrt on one side. Consequently you know the other side must be > 0. You also know that anything within the +sqrt > 0 (for real solutions). Fiddling around with these two equations will give you a domain in which your answers must lie. So if you get answers which lie outside your domain, you know they aren't solutions to your problem. No need to check your solutions

someonetoogoodforyou