How do you solve #2x^2 + 1 = 4x#?

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Answer 1 · 2015-08-07T03:13:10

#(1 + sqrt(2)/2)#, #(1 - sqrt(2)/2)#

#2x^2 + 1 = 4x#

Therefore: #2x^2 -4x +1 = 0#

If #ax^2 + bx + c = 0#

Then #x_1 = {-b + sqrt(b^2 - 4ac))/(2a)#

And #x_2 = {-b - sqrt(b^2 - 4ac)]/(2a)#

In this case #a=2#, #b=-4#, #c=1#

#x_1 = (-(-4) +sqrt((-4)^2-4 * (2) * (1)))/(2 * (2))#

#= (4 + sqrt(16-8))/4#

#= (4 + sqrt(8)]/4#

#=(4 + sqrt(2*2*2))/4#

#=(4 + 2sqrt(2))/4#

#= 1 + sqrt(2)/2#

Similarly:

#x_2 = 1 - sqrt(2)/2#