How do you solve #2x^2  x  5 = 0# by completing the square?
1 Answer
Explanation:
Premultiply by
#A^2B^2=(AB)(A+B)#
with
#0 = 8(2x^2x5)#
#color(white)(0) = 16x^28x40#
#color(white)(0) = (4x)^22(4x)+141#
#color(white)(0) = (4x1)^2(sqrt(41))^2#
#color(white)(0) = ((4x1)sqrt(41))((4x1)+sqrt(41))#
#color(white)(0) = (4x1sqrt(41))(4x1+sqrt(41))#
So:
#4x = 1+sqrt(41)#
and:
#x = 1/4+sqrt(41)/4#
Note that multiplying by

It makes the leading term into a perfect square ready for completing the square.

It makes the coefficient of
#x# into a multiple of twice the square root of the leading coefficient, thus avoiding having to use fractions until the end.