Question

$\frac{1}{(x-3)^2}- \frac{3}{x-3}-4=0$

Answer 1

$\frac{1}{(x-3)^2}- \frac{3}{x-3}-4=0 \\ \frac{1}{(x-3)(x-3)}- \frac{3}{x-3}- \frac{4}{1} =0 \\ \frac{1-3(x-3)-4(x-3)^{2}}{(x-3)(x-3)} = 0 \\ \frac{1-3x+9-4( x^{2} -6x+9)}{(x-3)(x-3)} = 0 \\ \frac{1-3x+9-4 x^{2} +24x-36}{(x-3)(x-3)} = 0 \\ \frac{-4 x^{2} +21x-26}{(x-3)(x-3)} = 0 \\ \left \{ {{{-4 x^{2} +21x-26=0} \atop {x-3 \neq 0}} \right. \\ \left \{ {{{-4 x^{2} +21x-26=0} \atop {x \neq 3}} \right.$
$\\ -4 x^{2} +21x-26=0 \\ D=21^{2}-4*(-26)*(-4)=441-416=25=5^{2} \\ x_{1}= \frac{-21- \sqrt{25}}{2*(-4)} = \frac{-21-5}{-8} = \frac{-26}{-8} =\frac{13}{4} =3 \frac{1}{4} =3.25 \\ x_{2}=\frac{-21+ \sqrt{25}}{2*(-4)} = \frac{-21+5}{-8} = \frac{-16}{-8} =2$