Radicals - Free Formula Sheet: bit.ly/48lpFLt Final Exams and Video Playlists: www.video-tutor.net/ Full-Length Math & Science Videos: www.patreon.com/mathsciencetutor/collections
I've started to watch your videos even though you've helped me PASS calculus and I no longer need math for my major. You're very calming and you're supurb at explaining the topic :))
8√5 = (2 x 4)√5. Take the 4 back inside the radical, which becomes 16, for 2√80. Now we're able to use a shortcut. 16 + 5 = 21 and 16 x 5 = 80. So automatically we have √16 - √5 which simplifies to 4 - √5. Whenever (a) you have 2 in front of the radical sign of the second term and (b) numbers that sum to the first term are (c) the same numbers of the product of the radicand of the second term, then you can use this shortcut. Keep the original sign, larger term placed first.
Actually this is hybrid operation involving factoring. The reduction of radicals is something like the 8th root of a is the 4th root of the square root etc
Your method only works for the cases you are using. As soon as the answer should have more than 1*the root your method fails badly. Try it on rt(37+20rt3). It will fail unless you realise you should be trying (a+brt3)^2 in you method. Here ab=10, and you have to try the different factors of 10 for a and b.
It works if you do it with simultaneous equations in a similar way he did with (A + B)^2. Let A^2 + B^2 = 37 be (p) and 2AB = 20√3 be (q) (q) implies that A = (10√3)/B which implies, through substitution into (p), that B^4 - 37B^2 + 300 = 0 which factories to (B^2 - 25)(B^2 - 12) = 0 so we can generate one combination where B = 5, which implies that A = 2√3 so the result Is √(37 + 20√3) = 5 + 2√3 which is correct.
Let √(37+20√3) = √x + √y Square both side: 37+20√3 = x+y+2√(xy) You have to manipulate 20√3 into 2√(something) 2•10√3 = 2√300 37+2√300 = (x+y)+(2√xy) x+y = 37 xy = 300 Solve it by substitution and you'll get x, y=12, 25 and 25, 12 √x + √y = √12 + √25 = 2√3 + 5 or 5 + 2√3
Radicals - Free Formula Sheet: bit.ly/48lpFLt
Final Exams and Video Playlists: www.video-tutor.net/
Full-Length Math & Science Videos: www.patreon.com/mathsciencetutor/collections
Can you believe me that I've spent the last 3 days trying to find someone explaining exactly this? I'm ever grateful for this video
Same situation lol, I've been trying to find this exact problem and i always find normal square root video instead
I've started to watch your videos even though you've helped me PASS calculus and I no longer need math for my major. You're very calming and you're supurb at explaining the topic :))
I agree with you!
You explain it better than my teacher. Thanks so much!
Can you make a tutorial on graphing calculators please?
8√5 = (2 x 4)√5. Take the 4 back inside the radical, which becomes 16, for 2√80. Now we're able to use a shortcut. 16 + 5 = 21 and 16 x 5 = 80. So automatically we have √16 - √5 which simplifies to 4 - √5. Whenever (a) you have 2 in front of the radical sign of the second term and (b) numbers that sum to the first term are (c) the same numbers of the product of the radicand of the second term, then you can use this shortcut. Keep the original sign, larger term placed first.
Impressive!
4:09 Answer = 3- radical 7 expand to 9 + 7 - 6 raidical 7 or 16 -6 radical 7
BRO... you explain better then my teacher!
Actually this is hybrid operation involving factoring. The reduction of radicals is something like the 8th root of a is the 4th root of the square root etc
I know you wrote this a year ago but it still is simplifying radical expressions
it doesn't work for all
me too
Thank you. Very very helpful. Could you tell me in which program you wrote. I am looking forward to hearing from you soon. Thank you
how would you do sqrt(10-2sqrt(5))
You should do an AMA on reddit
6:53 Answer =6+ radical 3
Your method only works for the cases you are using. As soon as the answer should have more than 1*the root your method fails badly. Try it on rt(37+20rt3). It will fail unless you realise you should be trying (a+brt3)^2 in you method. Here ab=10, and you have to try the different factors of 10 for a and b.
It works if you do it with simultaneous equations in a similar way he did with (A + B)^2.
Let A^2 + B^2 = 37 be (p)
and 2AB = 20√3 be (q)
(q) implies that A = (10√3)/B
which implies, through substitution into (p), that B^4 - 37B^2 + 300 = 0
which factories to (B^2 - 25)(B^2 - 12) = 0
so we can generate one combination where B = 5, which implies that A = 2√3
so the result Is √(37 + 20√3) = 5 + 2√3 which is correct.
Let √(37+20√3) = √x + √y
Square both side:
37+20√3 = x+y+2√(xy)
You have to manipulate 20√3 into 2√(something)
2•10√3 = 2√300
37+2√300 = (x+y)+(2√xy)
x+y = 37
xy = 300
Solve it by substitution and you'll get x, y=12, 25 and 25, 12
√x + √y = √12 + √25 =
2√3 + 5
or
5 + 2√3
Hi, what if i cant solve it by the method above?
Tnx ur video have been a lot of help keep going
Square root (a-b)^2 doesn’t equal a-b if a-b is a negative value bc of the principle square root property. You’d have to flip a-b to b-a.
Yes, I was going to say the same
sq root (a - b)^2 = (a-b) given that a > b
Thanks
Genius 🔥🔥
True! This. guy is amazing
Got all right sir, thanks for the help as I had thought in these problems!
شكرا لكم
I want to give people like a million dollars for helping students like me
Professor of science
hey bro pls can I know how you make your videos
I started watching you I became good in math
What about √6√3??
Thqnks but i couldn't get answer for my question
Kindful man
crazy work
How about in the cube root?
There's another method for that..
wao you did best job 😀🙂❤
great job
√5+4√6=?? Please answer
I don't see why the cross multiplication since you already started with the (A-B)^2=A^2+B^2-2AB
Love u broooo❤️❤️❤️❤️❤️
can someone explain me how to apply it in sqrt(21+4sqrt5) pleaseeeeeeee
still need help?
@@chair8401 yesss
Still need help？
AB = √(xy) = √20
A² + B² = x+y = 21
if xy = 20 and x+y = 21, x and y = 20 and 1
√(A² + 2AB + B²)
= A + B
= √20 + √1
= 2√5 + 1
I'm missing something obvious I'm sure. So √ a^2+b^2+2ab how does that = √(a+b)^2?:/( The last problem) thank you in advance!
Ok so foil but keep √?
yeah same question
love u!!!!!!
What if it’s Gallium
Bro a small but silly and important doubt if AB=4sq root of 5 then B=4÷Asq root of 5 right
There is an formula for solving these problems too. does anyone know that formula?
I have this one but under a few conditions:
a > 0 , b > 0 , and a^2 - b = k^2 (k > 0)
√a±√b = √(a + k)/2 ± √(a - k)/2
Nested radicals square roots
at first, it looks very simple problem but when you tried to solve it, its hard.
omg thank you
I understood that in 10 minutes but a teacher cant do it in 1 hiur
W teacher
Please answer my doubt please
Funny I'm learning radical exponents
i don't really get it at all
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