Question

Solve the equation for x, using factorization method

${x}^{2} + (y + \frac{1}{y}) + 1 = 0$
​

Answer 1

Appropriate Question

Solve for x, using factorization method

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

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

Given quadratic equation is

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

can be rewritten as

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

$\rm :\longmapsto\: {x}^{2} + xy + \dfrac{x}{y} + y \times \dfrac{1}{y} = 0$

$\rm :\longmapsto\: ({x}^{2} + xy) +\bigg( \dfrac{x}{y} + y \times \dfrac{1}{y}\bigg) = 0$

$\rm :\longmapsto\:x(x + y) + \dfrac{1}{y} \bigg(x + y\bigg) = 0$

$\rm :\longmapsto\:(x + y)\bigg(x + \dfrac{1}{y} \bigg) = 0$

$\rm :\longmapsto\:x + y= 0 \: \: \: or \: \: \: x + \dfrac{1}{y} = 0$

$\bf\implies \:x = - y \: \: \: or \: \: \: x = - \dfrac{1}{y}$

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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

Appropriate Question

Solve for x, using factorization method

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

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

Given quadratic equation is

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

can be rewritten as

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

$\rm :\longmapsto\: {x}^{2} + xy + \dfrac{x}{y} + y \times \dfrac{1}{y} = 0$

$\rm :\longmapsto\: ({x}^{2} + xy) +\bigg( \dfrac{x}{y} + y \times \dfrac{1}{y}\bigg) = 0$

$\rm :\longmapsto\:x(x + y) + \dfrac{1}{y} \bigg(x + y\bigg) = 0$

$\rm :\longmapsto\:(x + y)\bigg(x + \dfrac{1}{y} \bigg) = 0$

$\rm :\longmapsto\:x + y= 0 \: \: \: or \: \: \: x + \dfrac{1}{y} = 0$

$\bf\implies \:x = - y \: \: \: or \: \: \: x = - \dfrac{1}{y}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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