How do you solve $2 x^{2} + 1 = 4 x$ $2x^2 + 1 = 4x$ ?

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How do you solve $2 x^{2} + 1 = 4 x$ $2x^2 + 1 = 4x$ ?

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Answer 1 · 2015-08-07T03:13:10

$(1 + \frac{\sqrt{2}}{2})$ $(1 + sqrt(2)/2)$ , $(1 - \frac{\sqrt{2}}{2})$ $(1 - sqrt(2)/2)$

Explanation:

$2 x^{2} + 1 = 4 x$ $2x^2 + 1 = 4x$

Therefore: $2 x^{2} - 4 x + 1 = 0$ $2x^2 -4x +1 = 0$

Use quadratic formula:

If $a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$

Then $x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}$ $x_1 = {-b + sqrt(b^2 - 4ac))/(2a)$

And $x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}$ $x_2 = {-b - sqrt(b^2 - 4ac)]/(2a)$

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

$x_{1} = \frac{- (- 4) + \sqrt{{(- 4)}^{2} - 4 \cdot (2) \cdot (1)}}{2 \cdot (2)}$ $x_1 = (-(-4) +sqrt((-4)^2-4 * (2) * (1)))/(2 * (2))$

$= \frac{4 + \sqrt{16 - 8}}{4}$ $= (4 + sqrt(16-8))/4$

$= \frac{4 + \sqrt{8}}{4}$ $= (4 + sqrt(8)]/4$

$= \frac{4 + \sqrt{2 \cdot 2 \cdot 2}}{4}$ $=(4 + sqrt(2*2*2))/4$

$= \frac{4 + 2 \sqrt{2}}{4}$ $=(4 + 2sqrt(2))/4$

$= 1 + \frac{\sqrt{2}}{2}$ $= 1 + sqrt(2)/2$

Similarly:

$x_{2} = 1 - \frac{\sqrt{2}}{2}$ $x_2 = 1 - sqrt(2)/2$

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