# How do you solve by completing the square for x^2+ 3/2x = 3?

Mar 20, 2018

$x = + \sqrt{\frac{57}{16}} - \frac{3}{4} = 1.1374$

or

$x = - \sqrt{\frac{57}{16}} - \frac{3}{4} = - 2.6374$

#### Explanation:

${x}^{2} + \frac{3}{2} x = 3$

$\implies {x}^{2} + \frac{3}{2} x - 3 = 0$

$\implies {\left(x + \frac{3}{4}\right)}^{2} - {\left(\frac{3}{4}\right)}^{2} - 3 = 0$

$\implies {\left(x + \frac{3}{4}\right)}^{2} - \left(\frac{9}{16}\right) - 3 = 0$

$\implies {\left(x + \frac{3}{4}\right)}^{2} - \left(\frac{9}{16}\right) - \left(\frac{48}{16}\right) = 0$

$\implies {\left(x + \frac{3}{4}\right)}^{2} - \left(\frac{57}{16}\right) = 0$

$\implies {\left(x + \frac{3}{4}\right)}^{2} = \frac{57}{16}$

$\implies x + \frac{3}{4} = \pm \sqrt{\frac{57}{16}}$

$\implies x = \pm \sqrt{\frac{57}{16}} - \frac{3}{4}$

Therefore, x is either

$x = + \sqrt{\frac{57}{16}} - \frac{3}{4} = 1.1374$

$x = - \sqrt{\frac{57}{16}} - \frac{3}{4} = - 2.6374$

Mar 20, 2018

$x = \frac{- 3 \pm \sqrt{57}}{4}$

#### Explanation:

${x}^{2} + \frac{3}{2} x = 3$

Add ${\left(\frac{3}{4}\right)}^{2}$ both sides.

$\implies {x}^{2} + \frac{3}{2} x + {\left(\frac{3}{4}\right)}^{2} = 3 + {\left(\frac{3}{4}\right)}^{2}$

$\implies {\left(x\right)}^{2} + 2 \times x \times \frac{3}{4} + {\left(\frac{3}{4}\right)}^{2} = 3 + \frac{9}{16}$

$\implies {\left(x + \frac{3}{4}\right)}^{2} = \frac{57}{16}$

$\implies x + \frac{3}{4} = \frac{\pm \sqrt{57}}{4}$

$\implies x = \frac{\pm \sqrt{57}}{4} - \frac{3}{4}$

$\implies x = \frac{- 3 \pm \sqrt{57}}{4}$