Question

Solve for x

${x}^{2} - {y}^{2} - x - 3y - 2 = 0$
Explain every steps​

Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm \: {x}^{2} - {y}^{2} - x - 3y - 2 = 0$

can be re-arranged as

$\rm \: {x}^{2} - x- {y}^{2} - 3y - 2 = 0$

can be further rearrange as

$\rm \: {x}^{2} - x- ({y}^{2} + 3y + 2 )= 0$

$\rm \: {x}^{2} - x- [{y}^{2} + 2y + y + 2 ]= 0$

$\rm \: {x}^{2} - x- [y(y + 2)+ 1(y + 2) ]= 0$

$\rm \: {x}^{2} - x- (y + 2)(y + 1)= 0$

$\rm \: {x}^{2} + x( - 1)- (y + 2)(y + 1)= 0$

$\rm \: {x}^{2} + x( - 2 + 1)- (y + 2)(y + 1)= 0$

$\rm \: {x}^{2} + x( - 2 + 1 + y - y)- (y + 2)(y + 1)= 0$

$\rm \: {x}^{2} + x[(y + 1) - (y + 2)]- (y + 2)(y + 1)= 0$

$\rm \: {x}^{2} + x(y + 1) - x(y + 2)- (y + 2)(y + 1)= 0$

$\rm \: x(x + y + 1) - (y + 2)(x + y + 1) = 0$

$\rm \: (x + y + 1)[x - (y + 2)] = 0$

$\rm\implies \:x = - (y + 1) \: \: or \: \: x = y + 2$

Hence,

$\rm\implies \:\boxed{\sf{ \:\rm \: \: x = - (y + 1) \: \: or \: \: y + 2 \: \: }}$

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Additional Information

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

Answer 2

QUESTION :

• x - y - x - 3y - 2 = 0

SOLUTION :

• a x² + by² + 2g x 2 gy + C = 0

• a = 1 b = - 1

• a + b = 1 - 1 = 0

since a + b = 0. it represents pair of

parallel straight lines

• x ^ 2 - y ^ 2 - x + 3y - 2 = 0

• ( x - y +1) ( x + y - 2) = 0

• x + y x - 2 y

• - xy - y + 2 y

• x - y - x

• + 3y - 2

so, the answer will be :

x - y +1 = 0 or x + y - 2 = 0