Question

If ${\alpha }^2=5\alpha -3,{\beta }^2=5\beta -3$ then the value of $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$

• A. $19/3$
• B. $25/3$
• C. $-19/3$
• D. none of these

Answer 1

The correct option is A $19/3$

${\alpha }^2=5\alpha -3$ and ${\beta }^2=5\beta -3$

$\Rightarrow$ ${\alpha }^2-5\alpha +3=0$ and ${\beta }^2-5\beta +3=0$

$\Rightarrow$ $\alpha =\frac{5\pm \sqrt{25-12}}{2}$ and $\beta =\frac{5\pm \sqrt{25-12}}{2}$

$\Rightarrow$ $\alpha =\frac{5\pm \sqrt{13}}{2}$ and $\beta =\frac{5\pm \sqrt{13}}{2}$

Since, $\alpha \ne \beta$

So, take $\alpha =\frac{5+\sqrt{13}}{2}$ and $\beta =\frac{5-\sqrt{13}}{2}$

$\Rightarrow$ Now, $\frac{\alpha }{\beta }=\frac{\frac{5+\sqrt{13}}{2}}{\frac{5-\sqrt{13}}{2}}=\frac{5+\sqrt{13}}{5-\sqrt{13}}=\frac{(5+\sqrt{13}{)}^2}{(5{)}^2-(13{)}^2}$

$=\frac{25+13+10\sqrt{13}}{25-13}$

$=\frac{38+10\sqrt{13}}{12}$

$=\frac{19+5\sqrt{13}}{6}$ ---- ( 1 )

$\Rightarrow$ Now, $\frac{\beta }{\alpha }=\frac{\frac{5-\sqrt{13}}{2}}{\frac{5+\sqrt{13}}{2}}=\frac{5-\sqrt{13}}{5+\sqrt{13}}=\frac{(5-\sqrt{13}{)}^2}{(5{)}^2-(13{)}^2}$

$=\frac{25+13-10\sqrt{13}}{25-13}$

$=\frac{38-10\sqrt{13}}{12}$

$=\frac{19-5\sqrt{13}}{6}$ ----- ( 2 )

$\Rightarrow$ $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{19+5\sqrt{13}}{6}+\frac{19-5\sqrt{13}}{6}$ [ From ( 1 ) and ( 2 ) ]

$\Rightarrow$ $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{38}{6}=\frac{19}{3}$