How do you solve #q^ { 2} - 5q + 2= 0#?
1 Answer
Feb 23, 2017
Explanation:
Complete the square, then use the difference of squares identity:
#a^2-b^2 = (a-b)(a+b)#
with
#0 = 4(q^2-5q+2)#
#color(white)(0) = 4q^2-20q+8#
#color(white)(0) = (2q)^2-2(2q)(5)+25-17#
#color(white)(0) = (2q-5)^2-(sqrt(17))^2#
#color(white)(0) = ((2q-5)-sqrt(17))((2q-5)+sqrt(17))#
#color(white)(0) = (2q-5-sqrt(17))(2q-5+sqrt(17))#
#color(white)(0) = 4(q-5/2-sqrt(17)/2)(q-5/2+sqrt(17)/2)#
Hence:
#q = 5/2+-sqrt(17)/2#
