Test-1-1-1-1-1-1-i-i-i-2-1-so-1-1-Find-the-error-

Question Number 8600 by Chantria last updated on 17/Oct/16

T e s t

1 = \sqrt{1} = \sqrt{(- 1) (- 1)} = \sqrt{- 1} \cdot \sqrt{- 1}

= i \cdot i = i^{2} = - 1

s o 1 = - 1

F i n d t h e e r r o r .

Commented by prakash jain last updated on 17/Oct/16

{(a b)}^{x} = a^{x} \cdot b^{x}

only for a, b, x \in R and a, b, x are + ve .

{(- 1)}^{1 / 2} {(- 1)}^{1 / 2} \neq {(- 1 \times - 1)}^{1 / 2} since - 1 < 0

Answered by malwaan last updated on 18/Oct/16

\sqrt{1} = \pm 1

s o 1 = \sqrt{1} i s o n e p a r t f r o m t h e s o l u t i o n

a n d \sqrt{1} = - 1 i s t h e 2 n d p a r t

1 = \sqrt{1} = - 1 i s w r o n g

\sqrt{1} = \pm 1 i s r i g h t

Commented by Rasheed Soomro last updated on 19/Oct/16

“ \sqrt{}^{″} is used only for + ve root .

Hence \sqrt{1} = 1 not \sqrt{1} = \pm 1

We can write \pm \sqrt{1} = \pm 1 but

we {can}^{'} t write \sqrt{1} = \pm 1

Commented by FilupSmith last updated on 19/Oct/16

but \sqrt{x^{2}} = \pm x

{(+ x)}^{2} = x^{2}

{(- x)}^{2} = {(- 1)}^{2} {(x)}^{2} = x^{2}

Commented by nume1114 last updated on 19/Oct/16

\sqrt{x^{2}} =∣ x ∣

e x a m p l e

i f x = 3, \sqrt{x^{2}} = \sqrt{9} = 3, ∣ x ∣= 3

x = - 1, \sqrt{x^{2}} = \sqrt{1} = 1, ∣ x ∣= 1

x = 0, \sqrt{x^{2}} = \sqrt{0} = 0, ∣ x ∣= 0