How do you solve using the completing the square method #2x^2-9x-17=0#?

1 Answer
Mar 27, 2016

Answer:

#x = 9/4+-sqrt(217)/4#

Explanation:

In addition to completing the square, I will use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a = (4x-9)# and #b = sqrt(217)#.

To cut down on the need for fractions, multiply through by #2^3 = 8# first to get:

#0 = 16x^2-72x-136#

#=(4x-9)^2-81-136#

#=(4x-9)^2-217#

#=(4x-9)^2-(sqrt(217))^2#

#=((4x-9)-sqrt(217))((4x-9)+sqrt(217))#

#=(4x-9-sqrt(217))(4x-9+sqrt(217))#

#=16(x-(9+sqrt(217))/4)(x-(9-sqrt(217))/4)#

So #x = 9/4+-sqrt(217)/4#