Question

Solve for the following quadratic equation for x ; x^2 + ( a+b/a + a/a+b) x + 1 = 0​

Answer 1

Answer:

x= -a/a+b, or -(a+b)/a

Step-by-step explanation:

x^2+a+b/a x+a/a+b x+1=0

x(x+a/a+b )+a+b/a(x+1/a+b/a)=0

(x+a/a+b )(x+a+b/a)=0

x= -a/a+b, or -(a+b)/a

Answer 2

Question

$\large{ \sf{ {x}^{2} + ( \dfrac{a + b}{ a} + \dfrac{a}{a + b}) + 1 = 0 }}$

Solution

Given,

$\large{ \sf{ {x}^{2} + ( \dfrac{a + b}{ a} + \dfrac{a}{a + b}) + 1 = 0 }}$

The constant term can be expressed as :

$\sf{ \dfrac{a + b}{a} \times \dfrac{a}{ a + b} }$

$\implies \: \sf{x {}^{2} + \dfrac{a + b}{a} x + \dfrac{a}{a + b}x + \dfrac{a + b}{a}. \dfrac{a}{a + b} = 0 } \\ \\ \implies \: \sf{x(x + \dfrac{a + b}{a} ) + \dfrac{a}{a + b}(x + \dfrac{a + b}{a}) = 0} \\ \\ \implies \: \sf{(x + \dfrac{a}{a + b})(x + \dfrac{a + b}{a}) = 0} \\ \\ \large{\implies \: \boxed{ \boxed{\sf{x = - \frac{a}{a + b} \: or \: - \frac{a + b}{a} }}}}$