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Simplify the following. √(-128) √(-16) If possible, can I get a step-by-step?
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that's \[(i \sqrt{128} )(i \sqrt{16})\] i times i = i^2 = -1 \[-(\sqrt{128})(\sqrt{16})\] \[-\sqrt{(128)(16)}\] \[-4\sqrt{128}\] \[-4\sqrt{(64)(2)}\] \[-(4)(8)\sqrt{2}\] \[-32\sqrt{?}\]
this used the complex number property of -1 = i^2 \[\sqrt{-128} = \sqrt{64 \times i^2 \times 2}\] giving \[8i \sqrt{2}\] \[\sqrt{-16} = \sqrt{16 \times i^2} = 4i\]
Is this correct? √(-128) √(-16) =√(-1*128) √(-1*16) =√(-1)*√128 √(-1)*√16 =i√128 i√16 =i^2 √(128*16) =-√2048 =-√(1024*2) =-√1024*√2 =-32*√2 √(-128) √(-16)=32*√2
yes...only with -..you forgot that
ah, thank you. forgot the negative. -facepalms- lol
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