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There cannot be a real number solution to this question, because the two roots have different signs. If the both signs were positive, then the solution would be a positive number to give us a real number solution; and if the signs were both negative, then the solution will be a negative number to give us a real number solution. But because they are distinct signs (i.e. one positive, one negative) the solution must be an imaginary number, there's no chance of getting a real number solution. KABADOSH.

Let's solve:

√x + √(-x) = 2;
Square BS, since all the terms with roots are on the LHS

[√x + √(-x)]² = 2² = 4;

From the identity: (a+b)²=a² +2ab +b²
a = √x; b = √(-x)

[√x + √(-x)]² = (√x)² +2[√x][√(-x)] + [√(-x)]²;
Square cancels square-root
and √a × √b =√(a×b)

∴[√x + √(-x)]² = x +2√[x×(-x)] + (-x);
[√x + √(-x)]² = x +2√[-(x×x)] - x;
[√x + √(-x)]² = 2√(-x²) - x + x;
[√x + √(-x)]² = 2√(-x²);

REM [√x + √(-x)]² = 4;

[√x + √(-x)]² = 2√(-x²) = 4;
∴2√(-x²) = 4;
DIV BS by 2
2√(-x²) ÷ 2 = 4/2;
√(-x²) = 2;
Square BS
[√(-x²)]² = 2²;
Square cancels square-root

-x² = 4;
DIV BS by -1

x² = -4;
Square-root BS

x = ±√(-4);
x = ±√4 × √(-1); NOTE: √(-1) = i
x = ±2× i;
x = +2i, or -2i;
x = ±2i.

God bless BELLO VICTOR OJOFUJADU for his graphical contribution in the thumbnail design.

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