Thanks for this problem. There is an easier way to solve it. Multiply both sides of the equation by the sum of the two radicals. √(w+7) - √(w+2) = 1
(√(w+7) - √(w+2)) x (√(w+7) + √(w+2))= (√(w+7)+ √(w+2)) (w + 7) - (w + 2) = (√(w+7)+ √(w+2)) There are now two equations in w: (√(w+7) + √(w+2) = 5 (simplification of the last equation) √(w+7) -- √(w+2) = 1 (the original equation) Subtract the second equation from the first equation. 2 (√(w+2)) = 4 √(w+2) = 2 Square both sides w + 2 = 4 w = 2
Thanks for showing this Jim. My approach to solve this problem would be exactly the same way. Here we are cutting the time in half as we are avoiding the two steps that involves squaring both sides. Once the students get the hang of this method many of them probably will try this shortcut method first to see if there is easy solution! With more complex radical equation the beauty and the ease of this method is better realized. It is an awesome process!
@ Jim Vinci For this particular problem at least, I beg to differ if you’re saying that the instructor’s method is the inferior to the one you present here. Less chance of error and incomparably easier and faster his way. The instructor’s is the superior one between the two here. Why multiply and expand an already simple expression into a larger and more complex one. His method of squaring the radicals is much more efficient than the method you present. In effect and in principle the two methods are the same things and get you the same result, but your process introduces more opportunity to err and even more radicals to deal with unnecessarily. Good to know your method, but why go from point a to point b while going from left to right at as you go when the better, quicker, less error prone, and more efficient way to go is in a straight line. Squaring both sides of the equation right off the bat to clear radicals is far far the more logical, sensible, and quicker method than all the multiplying out a multitude of them to clear them as you do here. It’s one thing to know a variety of techniques and how to use them, but it’s quite another thing all together to know when and where to use and apply them.
Greetings. The solution to this expression is 2, W=2. We determine the value as follows. The square root of (W+7) minus the square root of (W+2)=1 can be rewritten as the square root of (W+7) =1 plus the square root of (W+2). Thereafter, we squared both sides to get (W+7)= 1+2 times the square root of (W+2) plus (W+2). Moving forward, we have W+7 -W-2-1=2 times the square root of (W+2), and 4= 2 times the square root of (W+2). That is, the square root of (W+2)=2 and squaring both sides gives W +2=4 from which it is determined that W=2. Lovely. The solution is tested and proven to be correct. The square root of of 9 minus the square root of 4 is equal to 1.
Ditto: a math textbook would make it legit by saying "By inspection, we find the solution is w=2." ;-) But then I also back-substituted to do the check even though I had forgotten the concept of extraneous roots. Old habits die hard.
lots of work? hell i did it in my head... i looked at it and noticed that if w=2 it would be sq. root of 9=3 minus the sq. root of 4=2 and 3-2=1. i did not even watch the video...
Solution by inspection is valid. However, the point of this video is to learn the METHOD of solving this type of problem so when you come across one you can't solve by inspection you'll know how to figure it out.
Thanks for this problem. There is an easier way to solve it. Multiply both sides of the equation by the sum of the two radicals.
√(w+7) - √(w+2) = 1
(√(w+7) - √(w+2)) x (√(w+7) + √(w+2))= (√(w+7)+ √(w+2))
(w + 7) - (w + 2) = (√(w+7)+ √(w+2))
There are now two equations in w:
(√(w+7) + √(w+2) = 5 (simplification of the last equation)
√(w+7) -- √(w+2) = 1 (the original equation)
Subtract the second equation from the first equation.
2 (√(w+2)) = 4
√(w+2) = 2
Square both sides
w + 2 = 4
w = 2
Thanks for showing this Jim. My approach to solve this problem would be exactly the same way. Here we are cutting the time in half as we are avoiding the two steps that involves squaring both sides. Once the students get the hang of this method many of them probably will try this shortcut method first to see if there is easy solution! With more complex radical equation the beauty and the ease of this method is better realized. It is an awesome process!
Hi Jim, wanna help w/ My homework? just teasing, but UR the go-to Guy.
@ Jim Vinci
For this particular problem at least, I beg to differ if you’re saying that the instructor’s method is the inferior to the one you present here. Less chance of error and incomparably easier and faster his way. The instructor’s is the superior one between the two here. Why multiply and expand an already simple expression into a larger and more complex one. His method of squaring the radicals is much more efficient than the method you present. In effect and in principle the two methods are the same things and get you the same result, but your process introduces more opportunity to err and even more radicals to deal with unnecessarily. Good to know your method, but why go from point a to point b while going from left to right at as you go when the better, quicker, less error prone, and more efficient way to go is in a straight line. Squaring both sides of the equation right off the bat to clear radicals is far far the more logical, sensible, and quicker method than all the multiplying out a multitude of them to clear them as you do here. It’s one thing to know a variety of techniques and how to use them, but it’s quite another thing all together to know when and where to use and apply them.
Very well explained tutorial! Thank you.
This was fun too. Thx John
Greetings. The solution to this expression is 2, W=2. We determine the value as follows. The square root of (W+7) minus the square root of
(W+2)=1 can be rewritten as the square root of (W+7) =1 plus the square root of (W+2). Thereafter, we squared both sides to get (W+7)=
1+2 times the square root of (W+2) plus (W+2). Moving forward, we have W+7 -W-2-1=2 times the square root of (W+2), and 4= 2 times the square root of (W+2). That is, the square root of (W+2)=2 and squaring both sides gives W +2=4 from which it is determined that W=2. Lovely.
The solution is tested and proven to be correct. The square root of of 9 minus the square root of 4 is equal to 1.
The derivative [half(1/sqrt(w+7)-half(swrt(w+2) ]is always -ve for w+2 ge 0. Hence at most one real solution. Hence w=2 is only solution.
No need for the apostrohe in "Lot's". Apostrophes usually indicate possession (not here) or deleted letters ("Lot's" = "lot is," not anything).
I stucked on 9:40 , why became +W +3? where they came from? please help me
w=2
√9 -√4 = 3-2 = 1
took one look, immediately got the solution w = 2
Ditto: a math textbook would make it legit by saying "By inspection, we find the solution is w=2." ;-) But then I also back-substituted to do the check even though I had forgotten the concept of extraneous roots. Old habits die hard.
guessed it in 1 min
1+ 1
lots of work? hell i did it in my head... i looked at it and noticed that if w=2 it would be sq. root of 9=3 minus the sq. root of 4=2 and 3-2=1. i did not even watch the video...
Solution by inspection is valid. However, the point of this video is to learn the METHOD of solving this type of problem so when you come across one you can't solve by inspection you'll know how to figure it out.
Not much work. I worked it out mentally andvw=2
why the annoying 'go ahead and... ' expression..