How do you solve #2x^2 + 1 = 4x#?
1 Answer
Explanation:
Therefore:
Use quadratic formula:
If
Then
And
In this case
#= (4 + sqrt(16-8))/4#
#= (4 + sqrt(8)]/4#
#=(4 + sqrt(2*2*2))/4#
#=(4 + 2sqrt(2))/4#
#= 1 + sqrt(2)/2#
Similarly:
Therefore:
Use quadratic formula:
If
Then
And
In this case
#= (4 + sqrt(16-8))/4#
#= (4 + sqrt(8)]/4#
#=(4 + sqrt(2*2*2))/4#
#=(4 + 2sqrt(2))/4#
#= 1 + sqrt(2)/2#
Similarly: