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If the value of $\sqrt{-5-12 i}+\sqrt{7+24 i}$ is a negative real number $k$, then $k=$
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The correct answer is:
$-6$
$Z=\sqrt{-5-12 i}+\sqrt{7+24 i}$
$\begin{array}{ll}\because & \sqrt{a+i b}= \pm\left(\sqrt{\frac{|Z|+a}{2}}+i \frac{b}{|b|} \sqrt{\frac{|Z|-a}{2}}\right) \\ \therefore & \sqrt{-5-12 i}= \pm\left(\sqrt{\frac{13-5}{2}}+i \frac{(-12)}{12} \sqrt{\frac{13+5}{2}}\right) \\ & = \pm(2-i \cdot 3) \\ \Rightarrow & \sqrt{7+24 i}= \pm\left(\sqrt{\frac{25+7}{2}}+\frac{i \cdot 24}{|24|} \sqrt{\frac{25-7}{2}}\right) \\ & = \pm(4+3 \cdot i) \\ \therefore & Z= \pm(2-3 i) \pm(4+3 i)= \pm 6 \\ \because & Z < 0 \Rightarrow Z=-6 .\end{array}$
$\begin{array}{ll}\because & \sqrt{a+i b}= \pm\left(\sqrt{\frac{|Z|+a}{2}}+i \frac{b}{|b|} \sqrt{\frac{|Z|-a}{2}}\right) \\ \therefore & \sqrt{-5-12 i}= \pm\left(\sqrt{\frac{13-5}{2}}+i \frac{(-12)}{12} \sqrt{\frac{13+5}{2}}\right) \\ & = \pm(2-i \cdot 3) \\ \Rightarrow & \sqrt{7+24 i}= \pm\left(\sqrt{\frac{25+7}{2}}+\frac{i \cdot 24}{|24|} \sqrt{\frac{25-7}{2}}\right) \\ & = \pm(4+3 \cdot i) \\ \therefore & Z= \pm(2-3 i) \pm(4+3 i)= \pm 6 \\ \because & Z < 0 \Rightarrow Z=-6 .\end{array}$
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