Question

Find the value of 'm ' for which x^2 + 3xy + x + my -m has

two linear factors in x and y , with integer coefficients.

Answer 1

concept :- ax² + by² + 2gx + 2fy + c represent a pair of straight lines , if and only if
abc + 2fgh - af² - bg² - ch² = 0


Here,pair of straight lines is x² + 3xy + 0.y² + x + my - m .
now, compare it with above General equation.
then, a = 1 ,
h = 3/2 ,
b = 0,
g = 1/2 ,
f = m/2 and c = - m

so, abc + 2fgh - af² - bg² - ch² = 1 × 0 × (-m) + 2 × (m/2) × 1/2 × 3/2 - 1 × (m/2)² - 0× (1/2)² - (-m)(3/2)² = 0

3m/4 - m²/4 + 9m/4 = 0
(m² - 12m)/4 = 0
m(m - 12) = 0
m = 0, 12

hence, for m = 0 and 12 , Given equation will be pair of straight lines.

case 1 :- when m = 0
x² + 3xy + x + 0.y - 0
x² + 3xy + x = x(x + 3y + 1)
= (1.x + 0.y + 0.1)(x + 3y + 1)

hence, m = 0 will be answer because you can see that , two linear factors in x and y with integer coefficients.

case 2 :- when m = 12
x² + 3xy + x +12y - 12 = x² + 4x - 3x - 12 + 3xy + 12y
= x(x + 4) - 3(x + 4) + 3y(x + 4)
= (x - 3 + 3y )(x + 4)
= (x + 3y - 3)(x + 4 )
= (x + 3y - 3)(1.x + 0.y + 4)

hence, m = 12 is also possible .

answers : m = 0, 12

Answer 2

Step-by-step explanation:

So m is 0,12

Hope this answer may help u mate