$$4 \sqrt{-81} = ?$$

Question

Accepted Answer

ধরি, $x=\sqrt[4]{-81} \Rightarrow x^4=-81 \Rightarrow x^4+81=$ 0 $$\Rightarrow x^4-(9 i)^2=0 \Rightarrow\left(x^2-9 iight)\left(x^2+9 iight)=0$$ অথবা, $$\begin{array}{l|l}	herefore \mathrm{x}^2=9 \mathrm{i} & \mathrm{x}^2=-9 \mathrm{i} \=\left(\frac{3}{\sqrt{2}}ight)^2+2 \cdot \frac{3}{\sqrt{2}} \cdot \frac{3}{\sqrt{2}} \mathrm{i} & =\left(\frac{3}{\sqrt{2}}ight)^2-2 \cdot \frac{3}{\sqrt{2}} \cdot \frac{3}{\sqrt{2}} \mathrm{i} \+\left(\frac{3}{\sqrt{2}} \mathrm{i}ight)^2 & +\left(\frac{3}{\sqrt{2}}ight)^2 \=\left(\frac{3}{\sqrt{2}}+\frac{3}{\sqrt{2}} \mathrm{i}ight)^2 & =\left(\frac{3}{\sqrt{2}}-\frac{3}{\sqrt{2}} \mathrm{i}ight)^2 \	herefore \mathrm{x}= \pm \frac{3}{\sqrt{2}}(1+\mathrm{i}) & 	herefore \mathrm{x}= \pm \frac{3}{\sqrt{2}}(1-\mathrm{i})\end{array}$$ অতএব, $\sqrt[4]{-81}= \pm \frac{3}{\sqrt{2}}(1 \pm \mathrm{i})$ মনে রাখার কৌশল $$\begin{aligned}& \sqrt[4]{-a^2}= \pm \sqrt{\frac{a}{2}}(1 \pm i) \& 	herefore \sqrt[4]{-81}= \pm \sqrt{\frac{9}{2}}(1 \pm i)= \pm \frac{3}{\sqrt{2}}(1 \pm i)\end{aligned}$$

Admission JU A প্রশ্নব্যাংক MCQ — \[4 \sqrt{-81} = ?\]

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