If λ1 and λ2(λ1λ2) are two values of λ for which the expression f(x, y)=x2+λ x y+y2-5 x-7 y+6 can be resolved as a product of two linear factors then find the value of 3 λ1+2 λ2.

Question

If λ1 and λ2(λ1>λ2) are two values of λ for which the expression f(x, y)=x2+λ x y+y2-5 x-7 y+6 can be resolved as a product of two linear factors then find the value of 3 λ1+2 λ2. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

a =1 ; h =(λ/2) ; b =1 ; g =-(5/2) ; f =-(7/2) ; c =6 D =0 ⇒ abc +2 fgh - af 2- bg 2- ch 2=0 6 λ2-35 λ+50=0 λ=(10/3), (5/2) text hence λ1=(10/3), λ2=(5/2) ⇒ 3 λ1+2 λ2=15

Q. If $λ_{1}$ and $λ_{2} (λ_{1} > λ_{2})$ are two values of $λ$ for which the expression $f (x, y) = x^{2} + λ x y + y^{2} - 5 x - 7 y + 6$ can be resolved as a product of two linear factors then find the value of $3 λ_{1} + 2 λ_{2}$ .

$a = 1; h = \frac{λ}{2}; b = 1; g = - \frac{5}{2}; f = - \frac{7}{2}; c = 6$
$D = 0 \Rightarrow ab c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$
$6 λ^{2} - 35 λ + 50 = 0$
$λ = \frac{10}{3}, \frac{5}{2} hence λ_{1} = \frac{10}{3}, λ_{2} = \frac{5}{2} \Rightarrow 3 λ_{1} + 2 λ_{2} = 15$

Q. If λ1​ and λ2​(λ1​>λ2​) are two values of λ for which the expression f(x,y)=x2+λxy+y2−5x−7y+6 can be resolved as a product of two linear factors then find the value of 3λ1​+2λ2​.

a=1;h=2λ​;b=1;g=−25​;f=−27​;c=6 D=0⇒abc+2fgh−af2−bg2−ch2=0 6λ2−35λ+50=0 λ=310​,25​ hence λ1​=310​,λ2​=25​⇒3λ1​+2λ2​=15

Q. If $λ_{1}$ and $λ_{2} (λ_{1} > λ_{2})$ are two values of $λ$ for which the expression $f (x, y) = x^{2} + λ x y + y^{2} - 5 x - 7 y + 6$ can be resolved as a product of two linear factors then find the value of $3 λ_{1} + 2 λ_{2}$ .

$a = 1; h = \frac{λ}{2}; b = 1; g = - \frac{5}{2}; f = - \frac{7}{2}; c = 6$
$D = 0 \Rightarrow ab c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$
$6 λ^{2} - 35 λ + 50 = 0$
$λ = \frac{10}{3}, \frac{5}{2} hence λ_{1} = \frac{10}{3}, λ_{2} = \frac{5}{2} \Rightarrow 3 λ_{1} + 2 λ_{2} = 15$