\[\sqrt{-1}\times\sqrt{-1}\] \[case (1)=>\sqrt{-1}\times\sqrt{-1}=\sqrt{-1\times-1}=\sqrt{1}=1\] \[case(2)=>\sqrt{-1}\times\sqrt{-1}=(\sqrt{-1})^2=-1\] which one is correct either first case or the second case can anyone say please calvin lin can you say it

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TopNewest\( \sqrt{a} \times \sqrt{b} = \sqrt{ab} \) if and only if \( a,b> 0 \) – Siddhartha Srivastava · 2 years, 5 months ago

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Case 2 is correct...

Since \(\sqrt{-1} = i\), then, the equation above is simply equal to \(i^{2} = -1\).. :) – Christian Daang · 2 years, 5 months ago

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Case 1 is incorrect since you cannot split the square-root if both radicands are negative. – B.S.Bharath Sai Guhan · 2 years, 5 months ago

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mr.subbu case 2 is correct – Saran .P.S · 1 year, 9 months ago

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