If the expression $6 x^{2} + 2 y^{2} + a x y - 5 x - 3 y + 1$ can be resolved into product of two linear factors, then $a$ takes the value :

2

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4

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8

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16

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Solution

The correct option is C $8$
We know that a second degree equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0 . . . . (1)$
represents a pair of straight lines if $a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0 . . . (2)$

Compare $(1)$ with $6 x^{2} + 2 y^{2} + A x y - 5 x - 3 y + 1 = 0$ we get

$a = 6, b = 2, h = \frac{A}{2}, g = - \frac{5}{2}, f = - \frac{3}{2}, c = 1$

Put the values of $a, b, c, f, g a n d h$ in $(2)$ we get

$12 + \frac{15}{4} A - \frac{54}{4} - \frac{50}{4} - \frac{m^{2}}{4} = 0$

$\Rightarrow 48 + 15 A - 54 - 50 - A^{2} = 0$

$\Rightarrow A^{2} - 15 A + 56 = 0$

$\Rightarrow (A - 7) (A - 8) = 0$

$\Rightarrow A = 7, 8$

But $A = 8$ satisfies the condition $(2)$

Therefore when $A = 8$ the expression $6 x^{2} + 2 y^{2} + A x y - 5 x - 3 y + 1 = 0$ can be resolved into two linear factors.

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