I think there is a simpler route allowing you to solve the problem in your head without pen and paper. Notice that there is an obvious trivial solution x=0. In step two 2*sqrt(x)=x divide by sqrt(x) and you get sqrt(x)=2. Square both sides to get x=4.
I love little simple videos like this tbh. The simplest way I see it is to just show that x is a number such that 2*sqrt(x) = sqrt(x)*sqrt(x) and then match terms. This immediately implies that sqrt(x) = 2
Be careful because when "cancelling" sqrt(x) you are actually dividing both terms by sqrt(x), of course you can't divide by 0... which means it is a separate case... which fits the equation.
@oahuhawaii2141 you're right. That was definitely oversight on my part, as 0 is a solution to the equation 2*sqrt(x) = x, I was tunnelling on getting the sqrt(x) = 2
@matheusjahnke8643 I actually didn't do the cancelling in my head, I just matched the terms to see you have 2 times sqrt(x) and sqrt(x) times sqrt(x), which effectively is cancelling but as in my deduction I'm looking at the fact that the other term attached to the sqrt(x) is a 2, which would make them equal. I suppose its that thinking that caused the oversight
sqrt(x) + sqrt(x) = x 2sqrt(x) = x 2 = x/sqrt(x) [x = sqrt(x) * sqrt(x)] 2 = sqrt(x)*sqrt(x)/sqrt(x) 2 = sqrt(x) 2 + 2 = x 4 = x Another way of coming to the solution, great video regardless.
Число 4 не может считаться полным решением, в данной подаче уравнения. В изначальном уравнении фигурируют знаки квадратного корня, а квадратный корень из 4 может быть отрицательным
The square root of 4 is ±2 (-2) + (-2) = -4 For the equation to work, the actual square root plus the actual square root has to equal x 0 × 0 = 0 0^0 = 0 0 + 0 = 0 0 - 0 = 0 X = 0
@higher_mathematics Please ignore the negative comments from the peanut gallery of self-absorbed parrots who likely have never done anything to advance the frontiers of mathematics, but act like your videos are beneath their proficiency level. I spent lots of years studying math and completely enjoy your channel, even when I might take a different approach. Don't like this channel, go start your own, maybe call it "Home of the 3rd Derivatives of Displacement" so you sound cool. Thank you Higher Mathematics!!
So funny are those who found the solution in 1 second :D. The real solution is a... debate as several people have noticed. What if sqrt(4) is -2? So the debate is that the equation was badly designed. It should be: abs(sqrt(x)) + abs(sqrt(x)) = x
To everyone complaining it’s not “Higher”, he wrote
E A SY in The title.
Study Math but study grammar too!
I think there is a simpler route allowing you to solve the problem in your head without pen and paper. Notice that there is an obvious trivial solution x=0. In step two 2*sqrt(x)=x divide by sqrt(x) and you get sqrt(x)=2. Square both sides to get x=4.
In math you have to write everything
2√x = x
(2√x)² = x²
4x = x²
x² - 4x = 0
x(x - 4) = 0
x = 0, x = 4
So X is a number whose square root is exactly half its value. 4 is the only number which has this property.
But you missed x = 0 .
I watch these just cause of his nice handwriting
Proof it
This one's easy but I appreciate your channel!
I love little simple videos like this tbh. The simplest way I see it is to just show that x is a number such that 2*sqrt(x) = sqrt(x)*sqrt(x) and then match terms. This immediately implies that sqrt(x) = 2
But you missed x = 0 .
Be careful because when "cancelling" sqrt(x) you are actually dividing both terms by sqrt(x), of course you can't divide by 0... which means it is a separate case... which fits the equation.
@oahuhawaii2141 you're right. That was definitely oversight on my part, as 0 is a solution to the equation 2*sqrt(x) = x, I was tunnelling on getting the sqrt(x) = 2
@matheusjahnke8643 I actually didn't do the cancelling in my head, I just matched the terms to see you have 2 times sqrt(x) and sqrt(x) times sqrt(x), which effectively is cancelling but as in my deduction I'm looking at the fact that the other term attached to the sqrt(x) is a 2, which would make them equal. I suppose its that thinking that caused the oversight
Higher mathematics ❌
Complexity mathematics✅
Nice job, Euler.
sqrt x + sqrt x = x
Square both sides.
x + x + 2x = x^2 [(a+b)^2 = a^2 + b^2 + 2ab]
2x + 2x = x^2
4x = x^2
4x - x^2 = 0
x(4 - x) = 0
4 - x = 0 [divide both sides by x]
4 = x. Ans.
sqrt(x) + sqrt(x) = x
2sqrt(x) = x
2 = x/sqrt(x)
[x = sqrt(x) * sqrt(x)]
2 = sqrt(x)*sqrt(x)/sqrt(x)
2 = sqrt(x)
2 + 2 = x
4 = x
Another way of coming to the solution, great video regardless.
Have a geometry problems?
I did it on my head
Also the answer is very easy to assume
√x+√x=x
=>2✓x=x
=>4x=x²
=>4x/x=x²/x
=>4=x
Please don't call it Higher Mathematics 😂
He wrote: “Fun and E A S Y” in The title.
He also say in the video, that it's easy, but some students are confused
Some people didn't develop reading skills
It is just simple. Not higher math but elementary math.
I don't know where i can use this stuff 😞
rad X + rad X = X or
2 • rad X = X or
4 • X = X² or
4 = X
X5 -1 =0, so x5 =1. Then x = 1 . Simple
2√x = x
Let t = √x
2t = t²
t² - 2t = 0
t(t - 2) = 0
t₁ = 0, t₂ = 2
x₁ = 0, x₂ = 4
When you say "Prove" don't you mean "Check"?
If there are more square roots of x, then x1 = 0 , x2 = (number of square roots of x) ²
Interesting😂😂😂😂😂😂😂😂😂😂😂
sqrtx + sqrtx = x
2(sqrtx) = x
2(sqrtx) = sqrtx*sqrtx
2 = sqrtx
2^2 = x
x = 4
√x+√x=x
2x=x
2x÷√x=x÷√x
2=√x
X=2^2=4 (This sum is not do hard. It is a class 8th-9th calculation in India😊😊😊
Число 4 не может считаться полным решением, в данной подаче уравнения. В изначальном уравнении фигурируют знаки квадратного корня, а квадратный корень из 4 может быть отрицательным
из квадратного корня никогда не выходит отрицательное число, это невозможно.
√4 = 2; никак не -2, открой учебник за 8-ой класс
Если икс в квадрате равен 4, то икс- это плюс минус два. Не нужно путать одно с другим. Не позорьтесь.
√x + √x = x
2√x = x
4x = x²
4 = x
Thats more simple
Best answer by far!
Very easy.
Sqrt(x)+Sqrt(x)=x x=0 x=4
I didn't even need to start the video. I immediately see that x = 4.
How you became at that level
But you missed x = 0 .
2*√x = x
√x*(√x - 2) = 0
√x = 0, 2
x = 0, 4
Take note that square root of 4 is plus or minus 2! Negative 4 is not equal to 4. How do you handle that?
If you plot y=2rootx and y=x. You will see that the solution is solid. x = -4 does not appear anywhere. Good question though.
I don’t have to think about it.
Similar question: solve x+x=x*x. Even easier 😉
x = 2 and x = 0
Preschool Olympiad ?
2 root x = x
(2 root x) squared = x squared
4x = x ^2
4= x^2/x
4 = x
The square root of 4 is ±2 (-2) + (-2) = -4 For the equation to work, the actual square root plus the actual square root has to equal x
0 × 0 = 0
0^0 = 0
0 + 0 = 0
0 - 0 = 0
X = 0
Before I watch an overstate solution, I say x = 4 and that's it.^^
But you missed x = 0 .
X=4
Your so called prove is not really a prove, but a check, to prove it I would use the properties of monotone functions
x 4 easy
4
x = 0, 4
You could have written the fourth step as
4x=x^2
4=x^2/x
4=x
x=4
But both are same🤷♂️
But this is even easier
It's a quadratic equation it always has 2 answers they may be same,real or non real
@@GigaNoobs-k1j yeah i know but we get the answer right?
@@praveen.m2591 you only got one of the 2 correct answers
@@GigaNoobs-k1j my bad
Thank you for watching. What do you think about this question? Have a great day and love math!❤❤❤
Very Good explanation
basics
can we do like this: 2=x/sqrt(x) -> ^2 -> 4=x^2/x -> x=4
@@dimak76 Zero is also a solution
@higher_mathematics Please ignore the negative comments from the peanut gallery of self-absorbed parrots who likely have never done anything to advance the frontiers of mathematics, but act like your videos are beneath their proficiency level. I spent lots of years studying math and completely enjoy your channel, even when I might take a different approach.
Don't like this channel, go start your own, maybe call it "Home of the 3rd Derivatives of Displacement" so you sound cool.
Thank you Higher Mathematics!!
x=0😊
x = 0, 4
So funny are those who found the solution in 1 second :D. The real solution is a... debate as several people have noticed. What if sqrt(4) is -2? So the debate is that the equation was badly designed. It should be: abs(sqrt(x)) + abs(sqrt(x)) = x
The real numbers has only one solution when u sqrt them. sqrt(4) can only be 2
When x=4, √x=√4=±2. When we take x=-2, then -2+-2=-4≠4. What do we say?
x=0
Answer is X=2; X=0
Don’t call it higher math
Bro in india students of age group 11-12 will solve 💪tougher question than this
Its obviously 0
But you missed x = 4 .
0;4
not gonna follow this anymore.
What the fuck it is
{x+x ➖ }=x^2 (x+x ➖}=x^2 {x ^2+x^2}= x^4 x2^2 (x ➖ 2x+2),
???
@@oahuhawaii2141 hello
😅4 моментально ❕
0?
x = 0, 4
X = 4
4
4
4