# If roots of 6x^2+3x+2=0 are a and b, what is the value of (1-a)/(1+a)+(1-b)/(1+b)?

Jan 31, 2018

$\frac{1 - a}{1 + a} + \frac{1 - b}{1 + b} = \frac{8}{5} = 1 \frac{3}{5}$

#### Explanation:

In an equation $p {x}^{2} + q x + r = 0$, if its roots are $a$ and $b$

then $a + b = - \frac{q}{p}$ and $a b = \frac{r}{p}$

As roots of $6 {x}^{2} + 3 x + 2 = 0$ are $a$ and $b$, we have

$a + b = - \frac{3}{6} = - \frac{1}{2}$ and $a b = \frac{2}{6} = \frac{1}{3}$

Hence $\frac{1 - a}{1 + a} + \frac{1 - b}{1 + b}$

= $\frac{\left(1 - a\right) \left(1 + b\right) + \left(1 - b\right) \left(1 + a\right)}{\left(1 + a\right) \left(1 + b\right)}$

= $\frac{1 - a + b - a b + 1 - b + a - a b}{1 + a + b + a b}$

= $\frac{2 - 2 a b}{1 + a + b + a b}$

= $\frac{2 - \frac{2}{3}}{1 - \frac{1}{2} + \frac{1}{3}}$

= $\frac{\frac{6 - 2}{3}}{\frac{6 - 3 + 2}{6}}$

= $\frac{\frac{4}{3}}{\frac{5}{6}}$

= $\frac{4}{3} \times \frac{6}{5} = \frac{8}{5} = 1 \frac{3}{5}$