If the expression $6 x^{2} + 2 y^{2} + 8 x y + b x - 3 y + 1$ can be resolved into product of two linear factors, then $b$ takes the value

5

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- 5

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3

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- 3

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Solution

The correct option is B $- 5$
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 $6 x^{2} + 2 y^{2} + 8 x y + B x - 3 y + 1 = 0$ we get

$a = 6, b = 2, h = 4, g = \frac{B}{2}, f = - \frac{3}{2}, c = 1$

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

$12 - 6 B - \frac{54}{4} - \frac{2 B^{2}}{4} - 16 = 0$

$\Rightarrow 48 - 24 B - 54 - 2 B^{2} - 64 = 0$

$\Rightarrow 2 B^{2} + 24 B + 70 = 0$

$\Rightarrow B^{2} + 12 B + 35 = 0$

$\Rightarrow (B + 5) (B + 7) = 0$

$\Rightarrow B = - 5, - 7$

But $B = - 5$ satisfies $(2)$

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

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