How do you solve #2x^2 - x - 5 = 0# by completing the square?

1 Answer
Dec 3, 2017

Answer:

#x = 1/4+-sqrt(41)/4#

Explanation:

Premultiply by #8# (to avoid some fractions), complete the square and use the difference of squares identity:

#A^2-B^2=(A-B)(A+B)#

with #A=(4x-1)# and #B=sqrt(41)# as follows:

#0 = 8(2x^2-x-5)#

#color(white)(0) = 16x^2-8x-40#

#color(white)(0) = (4x)^2-2(4x)+1-41#

#color(white)(0) = (4x-1)^2-(sqrt(41))^2#

#color(white)(0) = ((4x-1)-sqrt(41))((4x-1)+sqrt(41))#

#color(white)(0) = (4x-1-sqrt(41))(4x-1+sqrt(41))#

So:

#4x = 1+-sqrt(41)#

and:

#x = 1/4+-sqrt(41)/4#

Note that multiplying by #8# has a couple of effects:

  • 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.