How do you solve 2x^2 - 4x -1=0 by completing the square?

Nov 19, 2016

$x = 2.22 \text{ or } x = 0225$

Explanation:

$2 {x}^{2} - 4 x - 1 = 0$

Divide both sides by 2 to get just ${x}^{2}$

${x}^{2} - 2 x - \frac{1}{2} = 0 \text{ } \leftarrow$move $- \frac{1}{2}$ to the other side
${x}^{2} - 2 x \text{ "= 1/2" } \leftarrow$ add $\textcolor{red}{{\left(- \frac{2}{2}\right)}^{2}}$ to both sides
${x}^{2} - 2 x \text{ } \textcolor{red}{+ 1} = \frac{1}{2} \textcolor{red}{+ 1}$

${\left(x - 1\right)}^{2} \text{ "=3/2" } \leftarrow$ write as square of a binomial

$x - 1 \text{ "=+-sqrt(3/2)" } \leftarrow$ find square root of each side

$x = + \sqrt{\frac{3}{2}} + 1 \text{ "or x = -sqrt(3/2)+1" } \leftarrow$ solve for x

$x = 2.22 \text{ or } x = 0225$