Question

${\lambda }_1$ and ${\lambda }_2$ are two values of $\lambda$ such that $f(x,y)=x^2+\lambda xy+y^2-5x-7y+6$ can be resolute as product of $2$ linear functions & ${\lambda }_1>{\lambda }_2$. then find the value of $3{\lambda }_1+2{\lambda }_2$

Answer 1

Condition that, $f(x,y)=a{x}^2+2hxy+b{y}^2+2gx+2fy+c$ can be resolute as product of two linear functions is,

$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$

Here, $f(x,y)=x^2+\lambda xy+y^2-5x-7y+6$

$⟹\left|\begin{array}{ccc}1& \frac{\lambda }{2}& \frac{-5}{2}\\ \frac{\lambda }{2}& 1& \frac{-7}{2}\\ \frac{-5}{2}& \frac{-7}{2}& 6\end{array}\right|=0$

$⟹1(6-\frac{49}{4})-\frac{\lambda }{2}(3\lambda -\frac{35}{4})-\frac{5}{2}(\frac{-7\lambda }{4}+\frac{5}{2})=0$

$⟹\frac{-25}{4}-\frac{3{\lambda }^2}{2}+\frac{35\lambda }{8}+\frac{35\lambda }{8}-\frac{25}{4}=0$

$⟹-\frac{3{\lambda }^2}{2}+\frac{35\lambda }{4}-\frac{25}{2}=0$

$⟹3{\lambda }^2-35\lambda +50=0$

$⟹(2\lambda -5)(3\lambda -10)=0$

$\therefore \lambda =\frac{5}{2}$ or $\lambda =\frac{10}{3}$

i.e., ${\lambda }_1=\frac{10}{3}$ and ${\lambda }_2=\frac{5}{2}$

$\therefore 3{\lambda }_1+2{\lambda }_2=3\left(\frac{10}{3}\right)+2\left(\frac{5}{2}\right)=10+5=15$ (Ans)