You can get the equation 2 sqrt(x) = 1 + sqrt(1 - x) of 5:19 in absolutely normal steps, just starting from beginning and isolating a radical, then squaring both sides.
This time you can just square and square both sides and things cancel out beautifully. The only solution I found is 16/25=0,64, but more important thing is the domain. You can easily see that x is greater or equal to 0 and 1-x as well, so x is smaller or equal to 1. But taking the biggest radical into account, you receive new points on the x axis and finally the domain is only from (sqrt(5)-1)/2 to 1.
It should be mentioned that the domain of the radical function on the LHS should be the interval [ (sqrt(5) - 1)/2, 1], and since the candidate solution x = 16/25 lies there it is a valid solution.
It remains to check that x = 16/25 satisfies the other condition sqrt (1-x) < x, which is not trivial, because for example 1/2 doesn't respect this inequality, in fact, 1/2 > 1/4. Luckily sqrt (9/25) = 3/5 < 16/25 and the condition is satisfied.
Nicely done with the first method. It seems more universal for this kind of problems, because if not the cancelations, squaring both sides wouldn't work that well. But still, I couldn't resist squaring both sides at the beginning and see what I was going to get. : )
x= 0 IS a valid solution. You just have to check all possible signs from all of those radicals. sqrt(x) - sqrt(x - sqrt(1-x)) = 1 x = 0 sqrt(0) - sqrt(0 - sqrt(1- 0)) = 1 sqrt(0) - sqrt(0 - sqrt(1)) = 1 0 - sqrt(0 +/- 1) = 1 0 - sqrt(1) = 1 || 0 - sqrt(-1) = 1 0 +/- 1 = 1 || 0 +/- i = 1 0 + 1 = 1 || 0 - 1 = 1 || 0 + i = 1 || 0 - i = 1 YES || NO || NO || NO A similar thing happens for x = 16/25 although you get 8 possible equations, only 1 of which is valid.
@@SyberMath if you remember you took it down to 5x = 4*sqrt ( x ) if we divide by 5*sqrt ( x ) on both sides you get : sqrt ( x ) = 4 / 5 and that means x = ( 4 / 5 )^2 = 16 / 25
I love your first method of solving this radical equation. Thank you sir.
Np
Won my heart😍!
This approach was worth it, Great approach to solve a radical equation like this. 😀
Glad you liked it! 🥰
assume x =sin(a)*sin(a), it will be easier to simplify : sin(a)-sqrt(sin(a)*sin(a)-cos(a))=1->2*sin(a)-cos(a)=1 ->sin(a)=4/5->x=14/25
You can get the equation 2 sqrt(x) = 1 + sqrt(1 - x) of 5:19 in absolutely normal steps, just starting from beginning and isolating a radical, then squaring both sides.
At 2:38 you can square both sides of (1-x)^0.5 < x and find that the domain of the problem is 0.618 < x < 1 (5^0.5 - 1)/2 is approximatly 0.618
Nice!
This time you can just square and square both sides and things cancel out beautifully. The only solution I found is 16/25=0,64, but more important thing is the domain. You can easily see that x is greater or equal to 0 and 1-x as well, so x is smaller or equal to 1. But taking the biggest radical into account, you receive new points on the x axis and finally the domain is only from (sqrt(5)-1)/2 to 1.
Nice!
Simple and clear presentation of the topics. Thanks .DrRahul Rohtak Haryana India
So nice of you, Dr. Rahul!
It should be mentioned that the domain of the radical function on the LHS should be the interval [ (sqrt(5) - 1)/2, 1], and since the candidate solution x = 16/25 lies there it is a valid solution.
Mann thats a nice problem and youre approach was insane !!!
Thanks, man!
It remains to check that x = 16/25 satisfies the other condition sqrt (1-x) < x, which is not trivial, because for example 1/2 doesn't respect this inequality, in fact, 1/2 > 1/4. Luckily sqrt (9/25) = 3/5 < 16/25 and the condition is satisfied.
Good observation!
Very nice the first method
Thanks for liking!
Great 👌 math tricks.thank you sir.
All the best
Interesting trick :)
I did not know about it,now i will
Second method is simple and more systematic
Yes!
Second method was more straight forward 👍🏽
Very good
Got it using the second method!
Nicely done with the first method. It seems more universal for this kind of problems, because if not the cancelations, squaring both sides wouldn't work that well. But still, I couldn't resist squaring both sides at the beginning and see what I was going to get. : )
Good point! Thanks!
That first method was amazing!!
When I saw that bunch of radicals I tried to trig sub but that didn't turn out well.
Thank you for the video!!!
You're welcome! Glad it helped!
Trig substitutions should always be avoided at all costs, because they make the problem much harder.
x= 0 IS a valid solution. You just have to check all possible signs from all of those radicals.
sqrt(x) - sqrt(x - sqrt(1-x)) = 1
x = 0
sqrt(0) - sqrt(0 - sqrt(1- 0)) = 1
sqrt(0) - sqrt(0 - sqrt(1)) = 1
0 - sqrt(0 +/- 1) = 1
0 - sqrt(1) = 1 || 0 - sqrt(-1) = 1
0 +/- 1 = 1 || 0 +/- i = 1
0 + 1 = 1 || 0 - 1 = 1 || 0 + i = 1 || 0 - i = 1
YES || NO || NO || NO
A similar thing happens for x = 16/25 although you get 8 possible equations, only 1 of which is valid.
x-√(1-x)=1-2√x+x
-√(1-x)=1-2√x
√(1-x)=2√x-1
1-x=4x-4√x+1
-x=4x-4√x
-4√x=-x-4x
-4√x=-5x
x/√x=-4/-5
√x=4/5
x=16/25
x / sqrt ( x ) = sqrt ( x ) for x > 0
How do we use that here?
@@SyberMath if you remember you took it down to 5x = 4*sqrt ( x )
if we divide by 5*sqrt ( x ) on both sides you get : sqrt ( x ) = 4 / 5
and that means x = ( 4 / 5 )^2 = 16 / 25
@@michaelempeigne3519 Oh I see! You're right. We don't have to square both sides. Good thinking 😁
Dang, I thought I saw gold in the thumbnail.
Gold would be sqrt(x) + (x - sqrt(1 - x - sqrt(x)))=1
That is cool!
😁
This equation relates to Pythagorean triple the famous 3-4-5
How?
I like.... Is the best
По-японски это называется
Нахера та хата!
Два возведения в квадрат,сокращения и готов ответ.
Interesting! 😁
Easy
Good!