College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3 - Page 50: 127

Answer

a and b must be non-negative if $\sqrt a . \sqrt b$ = $\sqrt ab$ Let us consider that a and b are negative: then, $\sqrt -a . \sqrt -b$ = -$\sqrt ab$ $\ne$ $\sqrt ab$

Work Step by Step

a and b must be non negative if $\sqrt a . \sqrt b$ = $\sqrt ab$ To prove this let us consider that a and b are negative. $\sqrt -a$ = $(\sqrt a) $ $\times$ i where i = $\sqrt -1$ which is imaginary same for b $\sqrt -b$ = $(\sqrt b) $ $\times$ i $\sqrt -a . \sqrt -b$ = $(\sqrt a) $ $\times$ i $\times $$(\sqrt b) $ $\times$ i = $(\sqrt a) $ $\times $$(\sqrt b) $ $\times$$ i^{2} $ as $ i^{2} $ = -1 then $(\sqrt a) $ $\times $$(\sqrt b) $ $\times$$ i^{2} $ = -$(\sqrt a) $ $\times $$(\sqrt b) $ = -$\sqrt ab$ $\ne$ $\sqrt ab$ so a and b cannot be negative, they must be non-negative
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