# How do you solve 2x^2-4x-3=0 by completing the square?

Jul 11, 2016

$x = 1 \pm \sqrt{\frac{5}{2}} .$

Taking, $\sqrt{\frac{5}{2}} \cong 1.581$, we get the soln. set $= \left\{2.581 , - 0.581\right\} .$

#### Explanation:

Given that, $2 {x}^{2} - 4 x - 3 = 0.$

$\therefore 2 {x}^{2} - 4 x = 3.$

To complete the square on $L . H . S .$, we need
Last Term $= {\left(M i \mathrm{dl} . T e r m\right)}^{2} / \left(4 \times F i r s t T e r m\right) = {\left(- 4 x\right)}^{2} / \left(4 \cdot 2 {x}^{2}\right) = \frac{16 {x}^{2}}{8 {x}^{2}} = 2$

Adding $2$ on both sides,

$2 {x}^{2} - 4 x + 2 = 3 + 2$

$\therefore 2 \left({x}^{2} - 2 x + 1\right) = 5$

$\therefore 2 {\left(x - 1\right)}^{2} = 5$

$\therefore {\left(x - 1\right)}^{2} = \frac{5}{2}$

$\therefore \left(x - 1\right) = \pm \sqrt{\frac{5}{2}}$

$\therefore x = 1 \pm \sqrt{\frac{5}{2}} ,$ is the desired soln.

Taking, $\sqrt{\frac{5}{2}} \cong 1.581$, we get the soln. set $= \left\{2.581 , - 0.581\right\} .$