How do you solve the equation and identify any extraneous solutions for #(x-4)/x + x/3=6#?

Question

1637 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-17T18:06:21

Multiply through by #x# to get a quadratic in #x#, then use quadratic formula to get:

#x = (15+-sqrt(273))/2#

Multiply all the terms by #3x# to get:

#3(x-4)+x^2 = 18x#

In theory this could introduce an extraneous #x=0# solution, but it does not...

Subtract #18x# from both sides to get:

#x^2-15x-12 = 0#

This quadratic is in the standard #ax^2+bx+c# form, with #a=1#, #b=-15# and #c=-12#

Solve using the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#=(15+-sqrt(15^2-(4xx1xx-12)))/(2xx1)#

#=(15+-sqrt(225+48))/2#

#=(15+-sqrt(273))/2#