How do you solve the equation #2x^2-3x+1=0# by completing the square?
1 Answer
Jan 8, 2017
Explanation:
#f(x) = 2x^2-3x+1#
The difference of squares identity can be written:
#a^2-b^2 = (a-b)(a+b)#
We will use this with
First pre-multiply by
#0 = 8f(x)#
#color(white)(0) = 8(2x^2-3x+1)#
#color(white)(0) = 16x^2-24x+8#
#color(white)(0) = (4x)^2-2(4x)(3)+9-1#
#color(white)(0) = (4x-3)^2-1^2#
#color(white)(0) = ((4x-3)-1)((4x-3)+1)#
#color(white)(0) = (4x-4)(4x-2)#
#color(white)(0) = (4(x-1))(2(2x-1))#
#color(white)(0) = 8(x-1)(2x-1)#
So