√{a ± b√c} (a,b, &c being rationals) can be simplified to √y±√z iff {a² - b²c}= d² & d is rational. Then take y=(a+d)/2 and z = (a-d)/2. Here a=7, b=1, and c=13. ∴ d= √{7² - 1².13}= √{49 - 13}= √36= 6. So y= {7+6}/2 =13/2 & z={7 - 6}/2=1/2. ∴ √{7 -√13} = √(13/2) - √(1/2) = {√26 - √2}/2.
You are doing too much work with your first method. At 2:25 you have a + b = 7 4ab = 13 This means that (a − b)² = (a + b)² − 4ab = 49 − 13 = 36 So we have either a − b = 6 or a − b = −6. But we need only one of these values because inverting the sign of a − b amounts to swapping the values of a and b. So, let's take a − b = 6, then we imediately have a = ((a + b) + (a − b))/2 = 13/2 b = ((a + b) − (a − b))/2 = 1/2 A third method starts by noting that √(7 + √13)√(7 − √13) = √(49 − 13) = √36 = 6. Now let √(7 + √13) + √(7 − √13) = A √(7 + √13) − √(7 − √13) = B Squaring both sides of both these equations all square roots drop out and we find A² = 14 + 2·6 = 26 and B² = 14 − 2·6 = 2, so √(7 + √13) + √(7 − √13) = √26 √(7 + √13) − √(7 − √13) = √2 Adding these identities we have 2√(7 + √13) = √26 + √2, so √(7 + √13) = ½√26 + ½√2
@@SyberMath Hold up! Someone else uses my (a+b)^2 - 4ab = (a-b)^2 trick? I figured that one out on my own, so it's amazing to see someone else using it.
√{a ± b√c} (a,b, &c being rationals) can be simplified to √y±√z iff {a² - b²c}= d²
& d is rational. Then take y=(a+d)/2 and z = (a-d)/2. Here a=7, b=1, and c=13.
∴ d= √{7² - 1².13}= √{49 - 13}= √36= 6. So y= {7+6}/2 =13/2 & z={7 - 6}/2=1/2.
∴ √{7 -√13} = √(13/2) - √(1/2) = {√26 - √2}/2.
the fact that 13 is a prime does most of the trick
I like to use the second method.
How to learn to do this exactly? Assuming values like a and b always confuse me where we learn that?
Where u get radical 17 🤔🤔🤔🤔🤔🤔🤔at 5:27
Just a mistake
He got it out of the typo pool.
Where did (sqrt13+sqrt1) all over 2 come from
(√26+√2)/2
(√(13) + 1) / √2 is the trivial one
1/√2 + √13/√2
√26/2+√(1/2)
The answer is ≈3.256
😂
Split 7 as 6.5 and 0.5
So it is root 6.5 + root 0.5 = 3.2566
You are doing too much work with your first method. At 2:25 you have
a + b = 7
4ab = 13
This means that
(a − b)² = (a + b)² − 4ab = 49 − 13 = 36
So we have either a − b = 6 or a − b = −6. But we need only one of these values because inverting the sign of a − b amounts to swapping the values of a and b. So, let's take a − b = 6, then we imediately have
a = ((a + b) + (a − b))/2 = 13/2
b = ((a + b) − (a − b))/2 = 1/2
A third method starts by noting that √(7 + √13)√(7 − √13) = √(49 − 13) = √36 = 6. Now let
√(7 + √13) + √(7 − √13) = A
√(7 + √13) − √(7 − √13) = B
Squaring both sides of both these equations all square roots drop out and we find A² = 14 + 2·6 = 26 and B² = 14 − 2·6 = 2, so
√(7 + √13) + √(7 − √13) = √26
√(7 + √13) − √(7 − √13) = √2
Adding these identities we have 2√(7 + √13) = √26 + √2, so
√(7 + √13) = ½√26 + ½√2
Nice!
@@SyberMath Hold up! Someone else uses my (a+b)^2 - 4ab = (a-b)^2 trick? I figured that one out on my own, so it's amazing to see someone else using it.