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  • If a < 0 and b < 0, then √a√b is equal toOption: 1 -√(|a||b|)Option:

If a < 0 and b < 0, then √a√b is equal to

Option: 1

-√(|a||b|)


Option: 2

√(|ab|)


Option: 3

i√(ab)


Option: 4

None of these


Answers (1)

best_answer

 

 

Iota and powers of Iota -

The square of any real number, whether it is positive or negative or zero is always non-negative, i.e. x2 ≥ 0 for all x ∈ R. 

Hence, the equation x2 + 1 = 0 is not satisfied for any real value of x or not solvable in real  number system..

Thus, the equation  x2 + 1 = 0 has imaginary solution. ‘Eular’ was the first mathematician to introduced the symbol i (read as ‘iota’).  The imaginary number i is defined as the square root of −1 .  

Hence, the equation

 

\\\mathrm{x^2+1=0}\\\mathrm{\Rightarrow \;\;x^2=-1}\\\mathrm{or,\;\;x=\pm\sqrt{-1}=\pm \mathit{i} }\\\mathrm{Equation,\;x^2+1=0\;\;has\;two\;solution,\;\;x=\textit{i}\;\;and\;\;x=-\textit{i}.}

 

\\\mathrm{\sqrt{-1}=i}\\\mathrm{\;\;\;\;\;\i^2=(\sqrt{-1})^2=-1}\\\\\mathrm{We\:can\:write\:the\:square\:root\:of\:any\:negative\:number\:as\:a\:multiple\:of\:i.}\\\mathrm{Consider\:the\:square\:root\:of\;\;-25}\\\\\mathrm{\sqrt{-25}=\sqrt{25\left(-1\right)}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;=\sqrt{25}\sqrt{-1}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;=5i}

 

 

 

a and b both are negative  

 

\\\mathrm{\sqrt{-a}\sqrt{-b}=\sqrt{(-1)a}\sqrt{(-1)b}}\\\mathrm{\Rightarrow \sqrt{-1}\sqrt{a}\sqrt{-1}\sqrt{b}=\textit{i}^2\sqrt{a}\sqrt{b}}\\\mathrm{\Rightarrow -1\sqrt{|a||b|}}

 

so the value is -√(|a||b|)

Correct option is (a)

Posted by

manish

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