@MichaelRothwell1

1 year ago

Nice problem and nice solution! Just a couple of points that need attention, though. First, when you square both sides of the equation, you risk adding extraneous solutions, so the potential solutions found need to be checked (though I realise this would be challenging to do in a short video). As it turns out, both x=½i and x=-½i are solutions to the given equation √x+√(-x)=1. But if the equation had instead been √x+√(-x)=-1, squaring would give the same result as squaring the original equation, so the same potential solutions would be found. But as these potential solutions satisfy √x+√(-x)=1, these would be extraneous solutions for the equation √x+√(-x)=-1, which therefore has no solutions. Second, when you write √x√(-x)=√[x(-x)]=√(-x²), you are applying the rule √a√b=√(ab), which is of course true for a≥0 and b≥0, but in general we may have √a√b=√(ab) or √a√b=-√(ab). The second case applies, famously, in the case a=b=-1, where √a√b=i²=-1, but √(ab)=√[(-1)²]=√1=1. This issue is easily fixed by leaving the expression in the form √x√(-x), which when squared, gives x(-x)=-x² as before. I should mention I am assuming the usual convention that the notation √x means the principal square root of x.

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