How to solve by quadratic formula: #(p-q)x+(2q)/x=(p+q)# ?
1 Answer
Explanation:
Given:
#(p-q)x + (2q)/x = (p+q)#
Multiply both sides of the equation by
#(p-q)x^2+2q = (p+q)x#
Subtract
#(p-q)x^2-(p+q)x+2q = 0#
This is in standard form:
#ax^2+bx+c = 0#
with
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)#