How do you solve # 1/(4 -x) = x/8#?

1 Answer
May 30, 2018

Answer:

Below

Explanation:

#1/(4-x)=x/8#

#1/(4-x)times8/8 = x/8times(4-x)/(4-x)#

#8/(8(4-x)) = (x(4-x))/(8(4-x))#

#8 = x(4-x)#

#8 = 4x-x^2#

#x^2-4x+8=0#

Using the quadratic formula:

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

#x = (4+-sqrt(16-4times1times8))/(2times1)#

#x = (4+-sqrt(-16))/2#

If you haven't learnt complex numbers yet, then you can ignore the following lines and just state that since the discriminant is less than 0, then there is no solution to the equation.

#x = (4+-sqrt(16times-1))/2#

#x = (4+-4i)/2#

#x = 2+-2i#

#x^2-4x+8=0#

#(x-(2-2i))(x-(2+2i))=0#

#(x-2+2i)(x-2-2i)=0#