Math Olympiad | A Nice Algebra Problem

[@the_portal.17]

Square both sides and x cancels out -x. Remaining is 2sqrt(-x^2)=16 sqrt(-x^2)=8 -x^2=64 x^2=-64 x=+-8i

[@imshiruba]

look this is my first time sharing my own way, so correct me if I'm wrong. sqrt(x) + sqrt(-x) = 4 sqrt(x) + i sqrt(x) = 4 sqrt(x) (1+i) = 4 sqrt(x) = 4/1+i x = ± 16/1+2i-1 x = ± 16/2i x = ± 8i

[@markdevries1535]

Think a bit quicker to square the left and right side of the initial equation

[@ЕвгенийКовригин-ь2з]

То

[@ЕвгенийКовригин-ь2з]

То от

[@nasrullahhusnan2289]

I work the problem in two ways: • direcly squaring the given equation • factoring LHS by taking sqrt(x) out by noting that sqrt(-x)=sqrt[x(-1)] =[sqrt(x)][sqrt(-1)] =i×sqrt(x) Both give the same answer as in the video, x=±8i. But I find difficulty upon checking whether x=±8i are the correct solutions. For x=8i: LHS=sqrt(8i)+sqrt(-8i) =[sqrt(8i)](1+i) How can we show that it equals 4? After rethinking for a while, what follows show how I solve the problem which the validity can be easily checked: sqrt(x)+sqrt(-x)=4 given equation As sqrt(-x)=sqrt(x)]sqrt(-1)] =i×sqrt(x) we can factor out sqrt(x) to get [sqrt(x)](1+i) =4 sqrt(x)=4/(1+i) (i) 6:04 On the other hand sqrt(x)=sqrt[(-x)(-1)] =[sqrt(-x)][sqrt(-1)] =i×sqrt(-x) We can factor out sqrt(-x) to get [sqrt(-x)](i+1)=4 sqrt(-x)=4/(i+1) (ii) Note that RHS of (i) is equal to that of (ii). Thus sqrt(x)=sqrt(-x) --> x=-x The given equation becomes 2sqrt(±x)=4 --> sqrt(±x)=2 (iii) ±x=4 From (iii) it can be easily seen that sqrt(x)+sqrt(-x)=2+2=4. I think (iii) is absurd. x=-x can only be if x=0. 2sqrt(±x)=0 not 4. Maybe the problem is supposedly a diophantine equation: sqrt(x)+sqrt(y)=4 with x=y=4 is the solution.

[@deltalima6703]

Since 8e^(i(5pi/2)) = x Is a valid solution, the answer in the video is obviously wrong.

[@rclrd1]

4 = √x + √(-x). Square both sides: 16 = x + 2√x√(-x) + (-x) = 2√(x^2)√(-1) = ±2xi x = ±8i

[@mircoceccarelli6689]

Esistenza dell'equazione : x >/= 0 & - x >/= 0 => x >/= 0 & x x = 0 => 0 + 0 = 0 = 4 impossibile => x^(1/2) + i x^(1/2) = 4 => ......................................... x^2 + 64 = x^2 - ( 8 i )^2 = 0 ( x - 8 i )( x + 8 i ) = 0 x = 8 i & x = - 8 i 😊👍👋

[@andreasproteus1465]

Verify the solution. Then again, in this case, verification is harder than the solution.

[@aidinsalehi6990]

easy way: (1+i)*sqrt(x)=4

[@OzanKaraayvaz]

Çok uzun bir çözüm

[@SheermohammadBalooch]

It's wrong

[@85833]

倒讚