How to solve by quadratic formula: #(p-q)x+(2q)/x=(p+q)# ?

1 Answer
George C.
Nov 21, 2017

#x=1" "# or #" "x = (2q)/(p-q)#

Explanation:

Given:

#(p-q)x + (2q)/x = (p+q)#

Multiply both sides of the equation by #x# to get:

#(p-q)x^2+2q = (p+q)x#

Subtract #(p+q)x# from both sides to get:

#(p-q)x^2-(p+q)x+2q = 0#

This is in standard form:

#ax^2+bx+c = 0#

with #a=(p-q)#, #b=-(p+q)# and #c=2q#

It has roots given by the quadratic formula:

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

#color(white)(x) = ((p+q)+-sqrt((-(p+q))^2-4(p-q)(2q)))/(2(p-q))#

#color(white)(x) = ((p+q)+-sqrt((p^2+2pq+q^2)-(8pq-8q^2)))/(2(p-q))#

#color(white)(x) = ((p+q)+-sqrt(p^2-6pq+9q^2))/(2(p-q))#

#color(white)(x) = ((p+q)+-sqrt((p-3q)^2))/(2(p-q))#

#color(white)(x) = ((p+q)+-(p-3q))/(2(p-q))#

That is:

#x = ((p+q)+(p-3q))/(2(p-q)) = (2p-2q)/(2p-2q) = 1#

or:

#x = ((p+q)-(p-3q))/(2(p-q)) = (4q)/(2(p-q)) = (2q)/(p-q)#

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