Question

The number of integral solutions of the equation $7\left(\begin{array}{ccc}y& +& \frac{1}{y}\end{array}\right)-2\left(\begin{array}{ccc}{y}^2& +& \frac{1}{y^2}\end{array}\right)=9$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

Consider,

$7\left(\begin{array}{ccc}y& +& \frac{1}{y}\end{array}\right)-2\left(\begin{array}{ccc}{y}^2& +& \frac{1}{y^2}\end{array}\right)=9$

$⟹$ $7\left(\begin{array}{ccc}y& +& \frac{1}{y}\end{array}\right)-2\left(\begin{array}{ccc}{y}^2& +& \frac{1}{y^2}+2-2\end{array}\right)-9=0$

$⟹$ $7\left(\begin{array}{ccc}y& +& \frac{1}{y}\end{array}\right)-2{\left(\begin{array}{ccc}y& +& \frac{1}{y}\end{array}\right)}^2+4-9=0$

$⟹$ $7\left(\begin{array}{ccc}y& +& \frac{1}{y}\end{array}\right)-2{\left(\begin{array}{ccc}y& +& \frac{1}{y}\end{array}\right)}^2-5=0$

Let $(y+\frac{1}{y})=a$

$⟹7a-2a^2-5=0$

$⟹2a^2-7a+5=0$

$⟹2a^2-2a-5a+5=0$

$⟹(2a-5)(a-1)=0$

$\therefore a=\frac{5}{2},$ $a=1$

$⟹(y+\frac{1}{y})=\frac{5}{2},$ $(y+\frac{1}{y})=1$

$⟹2y^2-5y+2=0,$ $y^2-y+1=0$

$⟹(2y-1)(y-2)=0,$ $y=\frac{1\pm \sqrt{1-4}}{2}$

$⟹y=\frac{1}{2},y=2$, $y=\frac{1\pm \sqrt{-3}}{2}$ (not real)

Only $2$ is the integer.

Hence, $2$ is the integral solution.