Question

Please simplify the following using factorization method

${x}^{2} + ( \frac{a}{a + b} + \frac{a + b}{a} ) + 1 = 0$
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Answer 1

Appropriate Question

Solve the following using factorization method :-

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

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

Given expression is

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

can be rewritten as

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

$\rm :\longmapsto\: {x}^{2} + \dfrac{a}{a + b}x + \dfrac{a + b}{a}x + \dfrac{a + b}{a} \times \dfrac{a}{a + b}= 0$

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

$\rm :\longmapsto\:x\bigg(x + \dfrac{a}{a + b}\bigg) + \dfrac{a + b}{a}\bigg(x + \dfrac{a}{a + b}\bigg) = 0$

$\rm :\longmapsto\:\bigg(x + \dfrac{a}{a + b}\bigg)\bigg(x + \dfrac{a + b}{a}\bigg) = 0$

$\rm\implies \:x \: = \: - \: \dfrac{a}{a + b} \: \: or \: \: \: x \: = \: - \: \dfrac{a + b}{a}$

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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

Answer 2

Answer:

Appropriate Question

Solve the following using factorization method :-

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

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

Given expression is

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

can be rewritten as

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

$\rm :\longmapsto\: {x}^{2} + \dfrac{a}{a + b}x + \dfrac{a + b}{a}x + \dfrac{a + b}{a} \times \dfrac{a}{a + b}= 0$

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

$\rm :\longmapsto\:x\bigg(x + \dfrac{a}{a + b}\bigg) + \dfrac{a + b}{a}\bigg(x + \dfrac{a}{a + b}\bigg) = 0$

$\rm :\longmapsto\:\bigg(x + \dfrac{a}{a + b}\bigg)\bigg(x + \dfrac{a + b}{a}\bigg) = 0$

$\rm\implies \:x \: = \: - \: \dfrac{a}{a + b} \: \: or \: \: \: x \: = \: - \: \dfrac{a + b}{a}$

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