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:

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

How do you solve 2x^2 + 1 = 4x2x^2 + 1 = 4x?