Can sqrt(x) + sqrt(-x) = 2？
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- Опубліковано 6 січ 2025
- In this math tutorial, we solve a tricky square root equation, sqrt(x) + sqrt(-x) = 2. This problem may seem daunting, but with the right techniques and strategies, we can find the solution. We start by exploring the properties of square roots and simplifying the equation using algebraic manipulation. We then introduce a clever substitution technique to turn the equation into a quadratic equation, which we can easily solve using the quadratic formula. This video is perfect for high school or college students who want to improve their algebra skills and tackle challenging square root equations. Watch the video and boost your math confidence today!
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#justalgebra
Hey man I just wanted to say that I remember being in 4th grade and watching some of your videos and being so amazed and wanting to learn more, now I do understand it and I have to thank you for sparking my love of math
@@bprpmathbasics 8th
@@bprpmathbasics but also I have a question, if you had a problem such as |(x+3)(x-5)| = -|(x+1)(x-6)|, the graph of them wouldn’t intersect, but would there be an imaginary solution or no solution to a problem like that?
@@bprpmathbasics You sparked my appreciation for math in 9th grade and now I'm majoring in math in college!
@@BrennaButcher No, you wouldn't find a solution at all. Absolute value functions are defined as the distance from zero, the origin of the graph, so you will only find positive answers as distance is positive only, therefore you would always have n=-n, which isn't true for nonzero values.
What would happen if you were to input a complex number is that you would have to use Pythagorean Theorem. Take |3+4i| for example. The answer would be 5 because you take 3^2+4^2=c^2, and solve for c.
I may be missing something as I'm only in algebra 2 and I'm not bprp, so take this with a grain of salt. I hope that answers your question, and keep on at it! I love math as well as you do.
@@ericcastro9044 so why not try to find a solution, isn’t that the same thing people said before we discovered complex numbers?
I extracted i from the √(-x) to turn it into i√x
2 = √x + i√x
Then factorized for ✓x
2=(√x)(1+i)
Divided both sides by 1+i
2/(1+i)=√x
Squared to get x
4/(1+i)²=x
Develop the binomial square
4/(1+2i+i²)=x
;;i² = 1
4/(1+2i-1)=x
Add similar terms together
4/2i=x
Reduced
2/i=x
;;i^(-1)=-i
-2i = x
you forgot to include the negative solution while squering but otherwise pretty cool
@@tox1678 i didn't know you need to include a negative solution when squaring, just when taking the even root.
I did the same , but in the end I thought that it's kinda strange that the question is symmetrical and the answer is not, so I just included the +2i too. Not sure how does it come from squaring.
upd: oh yeah it comes from the first step, not squaring, -i works there too.
You lost the second solution in your first step by extracting only i and not (-i) as well.
@@derwolf7810 Yeah I know)
Didnt want to double track in a youtube comment)
I really like that your channels and videos are keeping my love for math up to date. One thing I love about math is how everything connects up together, like a video game with no mistakes (other than some “game breaking” things, like 0 over 0…). Still, thank you for making videos. I really enjoy them :P
I solved this mentally as follows:
Squaring both sides of the original equation gives x + 2sqrt(-x^2) - x = 4, hence sqrt(-x^2) = 2.
Squaring both sides of the last reduced equation gives -x^2 = 4, so x^2 = - 4,
therefore x= sqrt(-4) = (+/-) 2i.
Same
Same - I was wondering why he did it the hard way...
@@KailLabs Beats me, because he is usually quite smart.
Finally, a problem that I managed to solve! Loving these Just Algebra videos they’re fun to watch
5:21 plus or minus 2 *square root of i*
Sorry, that last statement at 5:50 is not correct. The square root of a complex number is not uniquely defined. Each square root has to solutions and you need to take that into account to validate that equation, so essentially there are 4 cases to consider when taking roots. If you don't, the imaginary terms may not cancel thus leading to a false statement. Apart from that, we ought to check our solutions anyway whenever we square both sides of an equation, shouldn't we?
Man it feals so good and satisfying to find a solution to this that was different from your own
Not sure why you moved things around so much. If you take the original form:
sqrt(x) + sqrt(-x) = 2
Square Both Sides
=> 2sqrt(-x^2) = 4
=> |ix| = 2
=> x = -/+ 2i
2:38 why not just squaring both sides of the initial equation? You get something interesting:
x+ (-x)+2√[x*(-x)]=4
=>2√(-x²)=4 =>2ix=4
=>ix=2 =>x=-2i
By the way... Where did the other solution go?
You can't write √x √-x as √x(-x)
But that property doesn't just need one to be positive?
How did you pass the i to the other side multiplying? Are you wrong or can you explain me how to do it?
Oh I understood i(-i)=1
Oh your error was that ✓(-x^2) isn't ix. In fact is i |x| then |x| = minus 2i ==> x = plus/minus 2i
Different approach for the rewriting:
sqrt(x) + sqrt(-x) = 2
sqrt(x) + sqrt(-1)sqrt(x) = 2 | square
x + 2×sqrt(x)×sqrt(-1)sqrt(x) - x = 4 | simplify
2xi = 4
xi = 2 => x = -2i (and x = +2i of course)
No need to sperate -1 from x
@@hassanawdi3793 of course there is no need. BPRP showed a way without doing it in his video.
However I prefer it to work with real sqrt only when usefull, so sqrt(-x) = isqrt(x).
Other users factored sqrt(x) + isqrt(x) instead of squarin it, as I did. All ways are totally fine. Just depends on what you prefer.
I did that too
xi = 2
hmm 🤔
5:23 I don't think you said what you meant...
I got to the -2i answer by factoring our the i, multiplying by sqrt(x), squaring, and dividing by 2xi. Then just moved the i to the top of the fraction which simplified to -2i.
The way I solved it was to split they sqrt(-x) into i*sqrt(x) and then factor out sqrt(x) to get (1+i)sqrt(x)=2, divide by 1+i, and square both sides to get x=-2i, although you also have to consider 2i due to it being an even function.
Yeah it's much easier that way
I did it that way as well, but could you explain how you get the positive 2i a little more?
wait, I just did the math like that, when I have sqrt(x) = 2/(1+i), then I square both sides, don't you get x = 4/i which converts to x = -4i? I just become more confused.
I did that, and I got x = +-2/i. Strangely, it works.
I got the same result!
5:17 shouldn't we take forgranted all solutions obtained by squaring on both sides ? Or does that only apply for real solutions ?
I don't think that the solutions obtained by squaring can be taken for granted, even when they are complex.
This particular question appears to be a special case, owing to the symmetry of the terms on the left hand side of the original equation.
@@davidellis1079 ok, thanks
couldn't it also be + - 2i^(any odd number) or do you not write that since it will just become + - switched?
oh yeah btw thats true but still unnescesarry
2=sqrtx+sqrt(-1*x)
2=sqrtx+isqrtx
2=(sqrtx)(1+i)
2/(1+i)=sqrtx
+-4/(1+i)^2=x
+-4/2i
+-2/i
Times up and down by i to rationalize
+-2i
In order to find a domain and define sqrt(x), we define it on a subset of the complex numbers and we must choose a suitable branch cut: I would choose the branch cut of negative real numbers
thanks for making these interesting videos
Square root is a multivalued function if extended to the complex numbers, so the result should depend on the branches, chosen for both sqrt(x) and sqrt(-x)
There is a more direct way:
2 = sqrt(x)+sqrt(-x)= sqrt(x) * (1 + i)
so sqrt(x) = 2/(1 + i) = 2/(1 + i) * (1 - i)/(1 - i) = 1 - i
so x = (1- i )² = -2i
because of symmetry of equation 2i has to be s solution as well.
and by sqrt being multivalued, sqrt(x)+sqrt(-x) is also -2, 2i and -2i
I simply just said that √(-x)=+/-i√x and then dealt with both of the cases simultaneously, but other than that I did the same way
That's what i initially thought of as well. Much straight forward in my opinion
@@NoNameAtAll2 i am not sure what you mean by that. I mean about the -2. If you mean that -2 is a solution, well it's not true, but i dont think that's what you meant.
Edit: I need to think about using a different verb than mean omg
@@unfall5521 Don't be so mean to yourself ;)
I still got lost following this. I may have to do some basic algebra review before watching this again. Still a great video!
I have strange feeling you broke every single law in calculating roots here
Oh, this was so cool, really weird how that turns out, but interesting of course
To prove that √x+√-x = 2 is not define in the real world I used another method:
(√x+√-x)²=2² -> x+2√x(-x) -x = 4
= 2√-x² = 4 -> √-x² = 2 -> -x²=2
= x² = -4, which is not defined in the real world
So you basically solved it completely and showed that the solutions are complex
That’s actually how I did it as well
Nice Video!
I was actually able to solve it but with another approach :)
sqrt(x) + sqrt(-x) = 2
sqrt(x) + sqrt(-1)sqrt(x) = 2
sqrt(x) + isqrt(x) = 2
sqrt(x)(1 + i) = 2 | ^2
+-x(1 + 2i -1) = 4
+-2ix = 4 | :2
+-xi = 2 | *i
x = +-2i
±?
I agree this ± makes no sense in this place. See my similar solution in the newest post.
This doesn't make any sense, the complex sqrt function is multi valued
It doesn't, unless you consider CASES when taking square roots (which he doesn't)
Using complex analysis it is can be simple reply to do, can you do it sir?
wow I did use exponential to solve this and got the same answer but I must admit your method is much easier x)
2i and -2i
Wait, how did you magiically determine sqrt of 2i + sqrt of (-2i) = 2???
It's incredible that I've repeated the same steps 5 times and got 5 different results...
Caught in 4 k
5:23 sqrt of i
i di d not know you had this channel
Many people in the comments were wondering how to move
-1 outside of the square root as the value i, and still get
two solutions. This way the equation can be solved without
squaring both sides. Here is the trick of how to do it. :)
Solution:
sqrt(x) + sqrt(-x) = 2
sqrt(x) + sqrt( (-1) * x) = 2
sqrt(x) + sqrt( (±i)^2 * x) = 2
sqrt(x) + (±i) * sqrt(x) = 2
sqrt(x)(1 ± i) = 2 | ^2
x * (1 ± i)^2 = 4
x * (1 ±2i -1) = 4
x * (±2i) = 4 | / (±2i)
x = 4/(±2i)
x = ±2/i
x = ±2i/(i^2)
x = ±2i
I solved it differently:
√x + i√x = 2
(1+i) √x = 2
√x = 2 / (1+i)
x = (2 / (1+i))²
x = 4 / 2i
x = -2i
Since the equation is symmetrical, we have to consider both 2i and -2i as solutions.
I solved it exactly as you did, but I did think there were two solutions, so I didn't write down the last sentence. I was done at x = -2i, but what you said makes sense.
Why not factoring x^2 + 4 in (x + 2i)(x - 2i)?
I solved it by making x = r²e^(2Θi) with real r and Θ, r≥0. Hence, √x = re^(iΘ) and √-x = re^(-iΘ)
=> 2 = √x + √-x = r(e^(iΘ)+e^(-iΘ)) = 2rcosΘ
=> rcosΘ = 1
As this is an equation with 2 variables, there are infinite solutions. For example:
If Θ=π/4, r=√2 and one solution is x = r²e^(2Θi) = 2i
For Θ=-π/4, x=-2i
For Θ=π/3, r=2 and x=-2+i2√3
i remember getting recommended this same video of yours but it was in chinese!
Then uhh, I’m pretty sure it’s a different video, because he’s Taiwanese
@@kepler4192 i know, just felt like mentioning it
@@NightWorshiper sure
I just squared root(x) + root(-x) to get 2*root(x(-x)) = 2*root(-x^2) = 4. so root(-x^2) = 2. Therefore, -x^2 = 4.
I don't know why you rearranged and therefore had 2 binomial expansions.
When I substitute 2i into sqrt(x) + sqrt(-x), I get 2i as result instead of the expected 2. Can someone show me how I get 2 from sqrt(2i) + sqrt(-2i) ?
It would be better if we square both sides when the equation is √x+√-x=2
So when we open the square, it would give us, x-x+2√-x²=4..
Then √-x²=2.. then squaring both the sides again...
x²=-4, and here we get the our complex solutions
But i got 1 - i;
x^½ + x^½ × i = 2
x^½ ( 1 + i ) = 2
x^½ = 2/1+i
On conjugating,
x^½ = 1 - i.
I got a completely different answer:
sqrt(x) + sqrt(-x) = 2, note: sqrt(-1) is equal to i
sqrt(x) + i*sqrt(x) = 2
(1 + i)sqrt(x) = 2
sqrt(x) = 2/(1 + i)
x = 4/(1^2 + 1*i + i^2)
x = 4/(1 + i - 1)
x = 4/i
x = -4i
Another solution:
sqrt(x) + sqrt(-x) = 2, note: sqrt(-1) is equal to i
sqrt(x) + i*sqrt(x) = 2
(1 + i)sqrt(x) = 2
sqrt(x) = 2/(1 + i), note: (a + b)(a - b) = (a^2 - b^2)
sqrt(x) = 2(1 - i)/(1 + i)(1 - i)
sqrt(x) = 2(1 - i)/(1 - (-1))
sqrt(x) = 2(1 - i)/2
sqrt(x) = 1 - i
x = 1^2 + (1)(-i) + (-i)^2
x = 1 - i - 1
x = -i
???
Your first one:
Is 4/i equal to -4i? I don't remember the reciprocal of i being -i. I'm not sure how to finish that one though.
Second one:
x = 1^2 + (1)(-i) + (-i)^2
It looks like you forgot to multiply the middle term by 2, which would make it 1^2 + 2(1)(-i) + (-i)^2. That and when you square it to get rid of the sqrt, you should have a +/- on one side. That would give you the same answer sensei got.
Looking again, it appears you forgot the two on that middle term on the top one too.
(a + b)^2 = a^2 + 2ab + b^2
If you change that and put in the +/-, you end up with x = +/- 2/i. If you're right that the reciprocal of i is -i, then you get the same answer as sensei there too.
@@chitlitlah whoops, thank you
Why not square both sides right away?
Saraha ma3andich zhar la f tssahib la f zwaj walit tangoul tawahad maynawad tawahda Matahmal..... '' ''
Seen many of his videos, still can't understand why he carries a pokeball
It's the microphone.
Oh thanks for letting me know
I'm not satisfied with the result because if x = 2i and we put it in the expression sqrt(2i) + sqrt(-2i) = ±(1+i) ±(1-i) = ±2 or ±2i. So you have 4 different results depending on which square root you choose. You can solve it if you only consider square roots that have positive results so you state that sqrt(x)≥0 for every x because then sqrt(2i) + sqrt(-2i) = (1+i) + (1-i) = 2.
Beautiful
Hmm. I assumed the real part of rootx and root-x must add to 2 so I assume rootx = 1 + bi. Root-x must be 1 - bi. So x and -x are, in either order, (1-b^2) + 2bi and (1-b^2) - 2bi. But this did not work. :P
I did it as follows: (1+i)sqrt(x) = 2 => 2xi = 4 => x = -2i. So why did I lose the +2i solution?
I was wondering the same. I found a correct way of solving it like you. See my post at the top of the comments.
well actually √(2i)+√(-2i) gives you 2 but also -2, 2i and -2i soooo are you sure x = ±2i is a good answer ?
@@jorgelenny47 I was wrong my bad, there’s actually 3 other values
@@jorgelenny47 maybe you want some explanations : your computation is totally right but you missed one thing. The square root for complex numbers is not a function, so for example if you compute √i there’s actually two values :
√i = (e^(iπ/2+2iπn))^(1/2)
= e^(iπ/4+iπn) = e^(iπ/4)•e^(iπn)
= ( cos(π/4)+i sin(π/4) )( e^(iπ) )^n
= (1/√2 + i/√2)(-1)^n
and just take n=0 and n=1 and you have you two values.
That’s the same thing in the video, when you compute √(2i)+√(-2i), it actually gives you (1+i)(-1)^n + (1-i)(-1)^m, so 4 values instead of only one, that’s why i’m asking if it’s ok to give x=±2i as a right answer
@@rshawty Ahhh yeah the +2πn, I forgot about that you're right. I guess we can accept it as the answer on the main rotation, but it's true that it is kinda sketcy
@@jorgelenny47 hehee but I think I have my answer finally :
{2e^(iπ/2(8n+1)),2e^(iπ/2(8n-1)) : n∈ℤ}
Has anyone controlled it? Because I did, but I found 4 more roots to the equation, ±2/i; ±i/2. Please sb tell me what did I do wrong
I got 2/i. I first made the LHS sqrt(x)(1+i) then divide by (1+1) leaving sqrt(x)=2/(1+i). Square both sides gives x=4/(1+2i-1)=2/i. I put sqrt(2/i) + sqrt(-2/i) into wolfram alpha and it said that's 2.
Or generally:
If √x + √-x = a
Then x = ±a²i½
Actually super Cool.
Thank you teacher ❤️
Can someone help me understand something, please? In the beginning of the video, he says that sqrt(-x) = 1 if x = -1.
But then why does the following show that it in fact equals -1, not 1?
sqrt(-x) = sqrt(-1 * x)
sqrt(-1 * x) = sqrt(-1) * sqrt(x) = i * sqrt(x)
if x = -1, then i * sqrt(x) = i * sqrt(-1) = i * i = i^2 = -1
This shows that the answer of sqrt(-x) when x = -1 is -1.
Am I wrong?
Maybe because i is rarely defined as i=sqrt (-1) but as i²=-1 as it doesn't follow many of the sqrt rules
It’s been a long time since I have studied this stuff. But my dumb question is how the square roots of 2i and -2i sum to 2? BPRP at the end kind of waves his hands and SAYS the sum is 2, but doesn’t show HOW they simplify to 2.
I got the answers +2i, -2i, 2/i. But everytime I try to check if the solution work I get something weird like 0=4
Is right to say that x=|-1|?
Got me fair and square. I got plus/minus root 2i. Solved it correctly graphically but forgot the root signs were already there. Lol
I did:
Sqrt(x) + isqrt(x) =2
Sqrt(x)(1+i)=2
Sqrt(x)=2/(1+i)
X=4/(1+2i-1)
X=4/2i
X=2/i
X=-2i
Which is only one out of the two answers, can anyone tell me why?
@@bprpmathbasics oh that makes sense thanks for the answer
then wouldn't the answer be +/- 1 ?
Nice
Your solution is overcomplicated and ignores that with principle square roots, there is no solution. There is no need to square the two sides at all, let alone twice. Simply by algebra you get sqrt(x)(1+/- i) = 2. This leads directly to sqrt(x) = 1 +/- i. Then, x = +/- 2i. But using principle square roots, the unique result of the starting expression is then 2i. Only by using the principle square root of 2i, and the non principle square root of -2i do you get 2 as a result. Note also, the considering all possible sums of all square roots of 2i and -2i, you can get 2i, -2i, 2, and -2 as possible results. In my opinion, you drastically overcomplicate the problem and also ignore the whole question of principle square roots.
√-2
Easy to sweeten from the clip image
I will try:
Sqrt(x)+sqrt(-x)=2. Square both side :
Sqrt(x)^2+sqrt(-x)^2+2sqrt(x)sqrt(-x)=4
x-x+ sqrt (x)*sqrt (-x)=sqrt(-x^2)=2. Square both side :
-x^2=4
x^2=-4
x= +sqrt(-4)=I* sqrt (4)= 2i or -sqrt (-4)=-i*sqrt (4)= -2i
Do you know why sqrt(-x^2) = sqrt(4) gives +- 2/i and not +- 2i , if -x^2 = 4 is equal to x^2 = -4. Help someone D:
√15
Yo quisiera comprobar que √2i + √-2i = 2, eso sí me tiene metido xD
√2i + √-2i = 2
√2√i + i√2√i = 2
√i√2( 1+ i) = 2
√i( 1+ i) = √2
i (1 + 2i - 1) = 2
-2 = 2 lol
Se puede demostrar que:
√2i = i + 1
√-2i = 1 - i
Si lo reemplazas en la ecuación te da 2
Sin embrago no sé que error cometí arriba para que me de -2=2 xDd
Bugué las matemáticas 🧐😎
Soy todo un admin
@@healthyhater un truco que puedes intentar es obtener el valor de √i. Hay dos valores posibles: √2/2 + i√2/2 o -√2/2 - i√2/2 (hay videos que muestran como obtener estos valores).
Entonces tienes 4 valores posibles para la ecuación √x + √-x:
a) si x = 2i y √i = √2/2 + i√2/2, entonces √x + √-x = 2i
b) si x = 2i y √i = -√2/2 - i√2/2, entonces √x + √-x = -2i
c) si x = -2i y √i = √2/2 + i√2/2, entonces √x + √-x = -2
d) si x = -2i y √i = -√2/2 - i√2/2, entonces √x + √-x = 2
De alguna manera en tu cálculo, me parece que te limitas a las soluciones a y b que son imaginarias puras. Al deshacerte de la parte imaginaria, la parte real no tiene sentido porque no existe.
Plugged into wolfram alpha and I still got answers I also did the math myself and also got an answer
I got easier solution:
Sqrt (x) * (1+i) = 2
I see that (1-i)(1+i) = 2, so
Sqrt (x) = 1 - i
Now square both sides and get
x = +- 2i
DONE!!!
I'm 8 grader so please make more these ytshort videos because they are cool and easy to solve in seconds
Square both sides then treat it like a quadratic
Damn I didn’t know I had to solve for complex numbers as well
Also uhh, to get all the answers, we can convert 2i to 2e^π(1/2i+2n) whereas n is an integer
If (1/2i + 2n) is part of the exponent, then you need additional grouping symbols.
5:21 + or - 2√i?? Lmao
haha I just took the minus out of the sqrt(-x) and turned it into :i*sqrt(x) and then factored with sqrt(x) which gave me the following:
sqrt(x)=2/1+i which equals sqrt(x)=1-i which also implies that x = +/-2i
答え
-2iだけじゃないですか？
I got 2/i.. it took me 10 mins to find out that i = 1/i. Wow
√1 + √-(-1) = 2
Vx+v-×=2 donc vx=2 car v-× it's impossible. Then it's impossible
But I think that's completely wrong bro ! No way
How about x=√(1) 😏
2i
Pitagoras...
You wrong at something (√(-x))^2 = |-x|=x. Don't remember like this and this will be cancel. Sometime it goes wrong
It’s the other way around. Sqrt(x^2)=abs(x)
But (sqrt(x))^2=x bc of complex numbers.
bprp seems a great guy, but the first two minutes of this video stink.