How do you solve 2x^2 + 1 = 4x2x2+1=4x?

1 Answer
Aug 7, 2015

(1 + sqrt(2)/2)(1+22), (1 - sqrt(2)/2)(122)

Explanation:

2x^2 + 1 = 4x2x2+1=4x

Therefore: 2x^2 -4x +1 = 02x24x+1=0

Use quadratic formula:

If ax^2 + bx + c = 0ax2+bx+c=0

Then x_1 = {-b + sqrt(b^2 - 4ac))/(2a)x1=b+b24ac2a

And x_2 = {-b - sqrt(b^2 - 4ac)]/(2a)x2=bb24ac2a

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

x_1 = (-(-4) +sqrt((-4)^2-4 * (2) * (1)))/(2 * (2))x1=(4)+(4)24(2)(1)2(2)

= (4 + sqrt(16-8))/4=4+1684

= (4 + sqrt(8)]/4=4+84

=(4 + sqrt(2*2*2))/4=4+2224

=(4 + 2sqrt(2))/4=4+224

= 1 + sqrt(2)/2=1+22

Similarly:

x_2 = 1 - sqrt(2)/2x2=122