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Question

Square root of x2+1x24i(x1x)6 where xεR is equal to :

A
±(x1x+2i)
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B
±(x1x2i)
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C
±(x+1x+2i)
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D
±(x+1x2i)
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Solution

The correct option is A ±(x1x+2i)

x2+1x24i(x1x)6

=x2+1x2+4ix4i1x42 (Using 1i=i)

=x2+1x2+4i2+2×x×2i2×1x×2i2×x×1x ----------(Using i2=1)

=(x)2+(1x)2+(2i)2+2×x(1x)+2×(1x)×2i+2×x×2i

=(x1x+2i)2 (Using (a+b+c)2=a2+b2+c2+2ab+2bc+2ac))

Therefore, square root of x2+1x24i(x1x)6 is ±(x1x+2i).

Hence, Option (A) is correct.


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