Chapter 6 - Section 6.5 - Solving Equations Containing Rational Expressions - Exercise Set - Page 379: 73

$-2$

$(\frac{3}{x-1})^{2}+2(\frac{3}{x-1})+1 = 0$ Let $u = (\frac{3}{x-1})$ Equation $(1)$ $u^{2}+2u+1=0$ It is in the form of perfect square trinomial. $[a^{2}+2ab+b^{2}= (a+b)^{2}$ $u^{2}+2u+1=(u+1)^{2}]$ $(u+1)^{2} = 0$ $(u+1) = 0$ $u= -1$ Substituting $u$ value in Equation $(1)$ $u = (\frac{3}{x-1})$ $-1= (\frac{3}{x-1})$ $-x+1 = 3$ $-x = 3 - 1$ $-x = 2$ $x = -2$