Question

If $\lambda_1$ and $\lambda_2\left(\lambda_1>\lambda_2\right)$ are two values of $\lambda$ for which the expression $f(x, y)=x^2+\lambda x y+y^2-5 x-7 y+6$ can be resolved as a product of two linear factors then find the value of $3 \lambda_1+2 \lambda_2$.

Answer 1

$a =1 ; h =\frac{\lambda}{2} ; b =1 ; g =-\frac{5}{2} ; f =-\frac{7}{2} ; c =6$
$D =0 \Rightarrow abc +2 fgh - af ^2- bg ^2- ch ^2=0$
$6 \lambda^2-35 \lambda+50=0 $
$\lambda=\frac{10}{3}, \frac{5}{2} \text { hence } \lambda_1=\frac{10}{3}, \lambda_2=\frac{5}{2} \Rightarrow 3 \lambda_1+2 \lambda_2=15$