Question

Find the square root of ${\left(\frac{a}{b}\right)}^2+{\left(\frac{b}{a}\right)}^2-\frac{4}{i}(\frac{a}{b}-\frac{b}{a})-6$

• A. $\pm \{(\frac{a}{b}+\frac{b}{a})-2i\}$
• B. $\pm \{(\frac{a}{b}-\frac{b}{a})-2i\}$
• C. $\pm \{(\frac{a}{b}+\frac{b}{a})+2i\}$
• D. $\pm \{(\frac{a}{b}-\frac{b}{a})+2i\}$

Answer 1

The correct option is D $\pm \{(\frac{a}{b}-\frac{b}{a})+2i\}$

${\left(\frac{a}{b}\right)}^2+{\left(\frac{b}{a}\right)}^2-\frac{4}{i}(\frac{a}{b}-\frac{b}{a})-6={\left(\frac{a}{b}\right)}^2+{\left(\frac{b}{a}\right)}^2+4i(\frac{a}{b}-\frac{b}{a})-6={\left(\frac{a}{b}\right)}^2+{(-\frac{b}{a})}^2+{\left(2i\right)}^2+2\left(\frac{a}{b}\right)(-\frac{b}{a})+2\left(\frac{a}{b}\right).2i+2(-\frac{b}{a}).2i={\{\pm (\frac{a}{b}-\frac{b}{a}+2i)\}}^{2}So,root:\pm (\frac{a}{b}-\frac{b}{a}+2i)$
Hence, (D) is correct.