I recognised the form of x+y±2√xy. This is (√x±√y)^2, same as the (a+b)^2 substitution. But you can use it to replace the roots directly. It's easier after multiplying by 2 (so the insides by 4). You get: √(√9+√7)^2+√(√9-√7)^2 = 2 a or a = ( 3+√7 + 3-√7 ) / 2 = 3
√{a ± b√c} (a,b, &c being rationals.) can be simplified to √y ± √z iff {a² - b².c} = d². If this d exists, then take y=(a+d)/2 and z = (a-d)/2. Here a = 7, b = 4, and c = 3. So d= √{4² - (½)².63}= √{16 - 63/4}= √{1/4} =½. ∴ y= (4+½)/2= 9/4 & z= {4-½}= 7/4. Thus √{4 + (½)√63} + √{4 - (½)√63} = (√y+√z)+(√y-√z)= 2.√y = 2√(9/4) = 2.(3/2) = 3.
I did it by squaring the whole thing, simplifying, and taking the square root: x = √(4+√63/2)+√(4-√63/2) x > 0 x^2 = 2*4 + 2*√(16-63/4) = 8 + √(64-63) = 9 x = 3
For me, an engineer rather than mathematician, the + on one half and the - on the other shouted ‘square it’ then I decided 63 was close to 64, so put in 64 to get root (8), to which I added root (1/64) to give 3 to 2 significant figures, which I decided was close enough for government work.
√8 + √(1/64) = 2*√2 + 1/8 = 2*1.414213562… + 0.125 = 2.953427124746… This misses 3 by about -1.5524%. Even if you truncate to 3 decimal places (√2 or last value), you get 2.953, which is off by about -1.5667%. Thus, your bridge collapsed -- many folks were killed, maimed, or injured.
Yes, I did it by squaring the whole thing, simplifying, and taking the square root: x = √(4+√63/2)+√(4-√63/2) x > 0 x^2 = 2*4 + 2*√(16-63/4) = 8 + √(64-63) = 9 x = 3
@@casparnigell5510 sqrt is defined to be >=0 for real numbers, that's also the reason why a>0, as both of the roots are positive (and at least one of them is non zero)
Sqrt(63) is slightly less than 8 and 8/2 = 4. So the second term is essentially 0. (Actually about 0.177). The first term is essentially sqrt(8) so 2sqrt(2) or about 2.828. Close enough for me.
I recognised the form of x+y±2√xy. This is (√x±√y)^2, same as the (a+b)^2 substitution.
But you can use it to replace the roots directly. It's easier after multiplying by 2 (so the insides by 4). You get:
√(√9+√7)^2+√(√9-√7)^2 = 2 a
or
a = ( 3+√7 + 3-√7 ) / 2 = 3
I worked out that the answer would be around 3, in about 10 seconds. Nice question, and precise answer by you.
Thank you!! Good excercise!🙂
This was good. Very gratifying!!
Thank you for these best questions 👍👍👍
Thanks bro, you doing great job
√{a ± b√c} (a,b, &c being rationals.) can be simplified to √y ± √z iff {a² - b².c} = d².
If this d exists, then take y=(a+d)/2 and z = (a-d)/2. Here a = 7, b = 4, and c = 3.
So d= √{4² - (½)².63}= √{16 - 63/4}= √{1/4} =½. ∴ y= (4+½)/2= 9/4 & z= {4-½}= 7/4.
Thus √{4 + (½)√63} + √{4 - (½)√63} = (√y+√z)+(√y-√z)= 2.√y = 2√(9/4) = 2.(3/2) = 3.
One of the easiest questions in this channel! Did it in 1 minute😋
it's not an olympiad mathematics. In my country it is for grade 9. Are you okay about your problem?
Really? Which country is it?
@@mathwindow Vietnam
người ta làm cho là tốt rồi còn vào cằn nhằn làm j
Hello 👋 madam please can I know the font type you use for your thumbnails
👍superb
I did it by squaring the whole thing, simplifying, and taking the square root:
x = √(4+√63/2)+√(4-√63/2)
x > 0
x^2 = 2*4 + 2*√(16-63/4)
= 8 + √(64-63)
= 9
x = 3
Show!!!👋👋👋👋
For me, an engineer rather than mathematician, the + on one half and the - on the other shouted ‘square it’ then I decided 63 was close to 64, so put in 64 to get root (8), to which I added root (1/64) to give 3 to 2 significant figures, which I decided was close enough for government work.
√8 + √(1/64)
= 2*√2 + 1/8
= 2*1.414213562… + 0.125
= 2.953427124746…
This misses 3 by about -1.5524%. Even if you truncate to 3 decimal places (√2 or last value), you get 2.953, which is off by about -1.5667%.
Thus, your bridge collapsed -- many folks were killed, maimed, or injured.
How did you approximate the 2nd square root √[4 - (√63)/2] with 1/64??
Спасибо.
fyi 1/2 is one half not one second
x^2+y^2=8,x.y=0.5
x+y=3
Yes, I did it by squaring the whole thing, simplifying, and taking the square root:
x = √(4+√63/2)+√(4-√63/2)
x > 0
x^2 = 2*4 + 2*√(16-63/4)
= 8 + √(64-63)
= 9
x = 3
But sqrt(1/4) is also -1/2. Why didn't you consider that 4 + 4 - 1= a^2 ?
Because a>0 so we only take one solution
@@aymencraftpro1878 Since both 9 and 7 are positive, I still don't understand why you took 9. As 7 is also > 0
@@casparnigell5510 sqrt is defined to be >=0 for real numbers, that's also the reason why a>0, as both of the roots are positive (and at least one of them is non zero)
the sqrt function notated √ is defined as to return only the principal root
I think this question is very much elementary
You should go for some hard and tricky question so that it will give a meaning to your channel
It can be solved more easily if we simplyfy this expression as sqrt(16+6sqrt(7)/4 ) + sqrt(16-6sqrt(7)/4)
31,5
Right answer is 3 but we got it by simplyfing this expression a lit bit another way
@@HK-dt2gf >:(
@@СвободныйМатематик Это ж банально) Всего навсего надо увидеть формулы квадрат суммы и кв разности
@@HK-dt2gf ага только коммент то отредачен был хех
شكرا
يمكن استعمال 63=9×7
3
Sqrt(63) is slightly less than 8 and 8/2 = 4. So the second term is essentially 0. (Actually about 0.177). The first term is essentially sqrt(8) so 2sqrt(2) or about 2.828. Close enough for me.
You're fine being shortchanged by 5.733% all the time? Tell your boss about that in your performance review and salary negotiations.
@@oahuhawaii2141 Brilliant. Irrelevant as hell, but brilliantly irrelevant.
@@jim2376: Yet your comment is the most irrelevant of this comments thread.
@@oahuhawaii2141 YOUR comment is the most irrelevant of this comments thread. But thanks for your bs anyway, you self important muppet.
without using a pen and no calculator I found it in less than 1min
I figured it out in my head in under 10 secs. Square the whole thing, simplify, and take the square root. Try it.