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If $y=\sqrt {x^2+6x+8}$, then the value of $\sqrt {1+iy}+\sqrt {1-iy}$= ?

$\begin{array}{1 1}(A)\sqrt {2(x+4)} \\ (B) \sqrt {(x+4)} \\ (C) \sqrt {2(x+3)} \\(D) \sqrt {(x+3)} \end{array}$

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Let $z=\sqrt {1+iy}+\sqrt{1-iy}$
$\Rightarrow\:z^2=1+iy+1-iy+2\sqrt{(1+iy)(1-iy)}$
$=2+2\sqrt{1+y^2}$
Given: $y=\sqrt {x^2+6x+8}$
$\therefore\:z^2=2+2\sqrt {1+x^2+6x+8}$
$=2+2\sqrt {(x+3)^2}=2(1+x+3)=2(x+4)$
$\Rightarrow\:z=\sqrt {2(x+4)}$
answered Jul 24, 2013 by rvidyagovindarajan_1
 

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