Question

Find x using factorization method

${x}^{2} + 3x - ( {y}^{2} + y - 2) = 0$
​

Answer 1

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

Given quadratic equation is

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

Here,

$\red{\rm :\longmapsto\: Constant \: term \: = \: {y}^{2} + y - 2}$

$\rm \:  =  \: {y}^{2} + 2y - y - 2$

$\rm \:  =  \: y(y + 2) - 1(y + 2)$

$\rm \:  =  \: (y + 2)(y - 1)$

Now,

$\red{\rm :\longmapsto\:Coefficient \: of \: x \: = \: 3}$

$\rm \:  =  \: 2 + 1$

$\rm \:  =  \: 2 + 1 + y - y$

$\rm \:  =  \: (y + 2) - (y - 1)$

So, given quadratic equation can be rewritten as

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

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

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

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

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

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