Question

solve qudrtic equation for x :
$x {}^{2} + ( \frac{a}{a + b} + \frac{a + b}{a} )x + 1 = 0$
​

Answer 1

HeYa❤️...

Answer:

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

I hOpe It HeLpS Ya!

✌️✌️✌️✌️

Answer 2

Answer:

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

$\begin{lgathered} \implies\sf{x}^{2} + \frac{a}{a + b} x + \frac{a + b}{a} x + 1 = 0\\ \\ \implies\sf x\bigg(x + \frac{a}{a + b} \bigg) + \frac{a + b}{a} \bigg(x + \frac{a}{a + b}\bigg)= 0\\ \\ \implies\sf\bigg(x + \frac{a}{a + b}\bigg)\bigg(x + \frac{a + b}{a}\bigg) = 0 \\ \\ \implies \boxed{\sf x = \frac{ - a}{a + b} \: \: or \: \:\frac{ - (a + b)}{a}}\end{lgathered}$