Imaginary!

Algebra Level 2

− 3 × − 4 − 6 × − 8 = ? \large\dfrac{\sqrt[]{-3} \times \sqrt[]{-4}}{\sqrt[]{-6} \times \sqrt[]{-8}} =\, ?

1 1 i i Undefined − 1 2 -\frac{1}{2} 1 2 \frac{1}{2} − 1 -1

Viki Zeta by Viki Zeta , 19, India

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2 solutions

Viki Zeta
Jul 9, 2016

− 3 × − 4 − 6 × − 8 = 3 i × 4 i 6 i × 8 i = ( 3 × 4 ) i 2 ( 6 × 8 ) i 2 = 3 × 4 6 × 8 = 12 48 = 12 48 = 1 4 = 1 2 \frac{\sqrt[]{-3} \times \sqrt[]{-4}}{\sqrt[]{-6} \times \sqrt[]{-8}}\\ = \frac{\sqrt[]{3}i \times \sqrt[]{4}i}{\sqrt[]{6}i \times \sqrt[]{8}i} \\ = \frac{(\sqrt[]{3} \times \sqrt[]{4})i^2}{(\sqrt[]{6} \times \sqrt[]{8})i^2} \\ = \frac{\sqrt[]{3} \times \sqrt[]{4}}{\sqrt[]{6} \times \sqrt[]{8}} \\ = \frac{\sqrt[]{12}}{\sqrt[]{48}} \\ = \sqrt[]{\frac{12}{48}} \\ = \sqrt[]{\frac{1}{4}} \\ = \frac{1}{2}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

root of 1/4 is plus-minus 1/2, and not only plus 1/2

Silver Vice - 4 years, 11 months ago

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This is root of imaginary units

− 3 2 × − 3 2 = 3 2 i × 3 2 i = 3 i 2 = − 3. ± a 2 = n , only applies for all a > 0 \sqrt[2]{-3} \times \sqrt[2]{-3} = \sqrt[2]{3}i \times \sqrt[2]{3}i = 3i^2 = -3. \\ \pm \sqrt[2]{a} = n, \text{ only applies for all a > 0}

Viki Zeta - 4 years, 11 months ago

− 3 × − 4 − 6 × − 8 \dfrac{\sqrt{-3} \times \sqrt{-4}}{\sqrt{-6} \times \sqrt{-8}}

= − 12 − 48 =\dfrac{\sqrt{-12}}{\sqrt{-48}}

= 4 ( − 3 ) 16 ( − 3 ) =\dfrac{\sqrt{4(-3)}}{\sqrt{16(-3)}}

= 2 − 3 4 − 3 =\dfrac{2\sqrt{-3}}{4\sqrt{-3}}

= 2 4 =\dfrac{2}{4}

= 1 2 =\boxed{\dfrac{1}{2}}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

First step is misleading − a = a i ∀ a > 0 − 3 × − 4 = 12 i 2 = − 12 \sqrt[]{-a} = \sqrt[]{a}i \forall a>0\\ \sqrt[]{-3} \times \sqrt[]{-4} = \sqrt[]{12}i^2=-\sqrt[]{12}

Viki Zeta - 3 years, 8 months ago

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