How do you solve \frac { 2} { ( 3k - 10) ^ { 2} } + \frac { 5} { 3k - 10} + 2= 02(3k10)2+53k10+2=0?

1 Answer
Nov 6, 2016

k=\frac{8}{3}or\frac{19}{6}k=83or196

Explanation:

multiply both sides by (3k-10)^2(3k10)2
2+5(3k-10)+2(3k-10)^2=02+5(3k10)+2(3k10)2=0
expand
2+15k-50+2(9k^2-60k+100)=02+15k50+2(9k260k+100)=0
15k-48+18k^2-120k+200=015k48+18k2120k+200=0
18k^2-105k+152=018k2105k+152=0
qudratic formula
k=\frac{105\pm\sqrt{(-105)^2-4(18)(152)}}{2(18)}k=105±(105)24(18)(152)2(18)
k=\frac{105\pm\sqrt{11025-10944}}{36}k=105±110251094436
k=\frac{105\pm\sqrt{81}}{36}k=105±8136
k=\frac{105\pm9}{36}k=105±936
k=\frac{35\pm3}{12}k=35±312
k=\frac{32}{12}or\frac{38}{12}k=3212or3812
k=\frac{8}{3}or\frac{19}{6}k=83or196