For a quadratic equation $a x^{2} + 2 x (h y + g) + b y^{2} + 2 f y + c = 0$ to be resolved into two linear factors, it must satisy which of the following conditions:

a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0

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a b + 2 f g h - a^{2} f - g^{2} b - h^{2} c = 0

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a b c + f g h - a f - b g - c h = 0

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a f^{2} - b g^{2} - c h^{2} = 0

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Solution

The correct option is A $a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$
Its basic fact that, $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ can be resolved into two linear factors if,
$a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$

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Similar questions

If the pair of lines $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ intersect on the y axis, then

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2}

(f^{2} - b c) x + (c h - f g) y + (b g + h f) z = 0

(c h - f g) x + (g^{2} - c a) y + (a f - g h) z = 0

(b g - h f) x + (a f - g h) y + (h^{2} - a b) z = 0

a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0

3 x^{2} - 2 y^{2} + 4 x - 6 y = 0

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} \neq 0

h^{2} > a b

(x + 1) (y - 3) = 4

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

a + b = 0, h^{2} > a b

a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0

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