How do you solve using the completing the square method x^2-5x+3=0?

Nov 4, 2017

$x = \frac{1}{2} \left(5 \pm \sqrt{13}\right)$
[$x \approx 4.30278 \mathmr{and} \approx 0.69722$]

Explanation:

${x}^{2} - 5 x + 3 = 0$

${x}^{2} - 5 x = - 3$

Our objective here is to make the LHS into a perfect square.

Add ${\left(\frac{5}{2}\right)}^{2}$ to both sides:

${x}^{2} - 5 x + {\left(\frac{5}{2}\right)}^{2} = - 3 + {\left(\frac{5}{2}\right)}^{2}$

Notice that the LHS $= {\left(x - \frac{5}{2}\right)}^{2}$

Hence, ${\left(x - \frac{5}{2}\right)}^{2} = - 3 + \frac{25}{4}$

${\left(x - \frac{5}{2}\right)}^{2} = \frac{- 12 + 25}{4} = \frac{13}{4}$

$\therefore x - \frac{5}{2} = \pm \sqrt{\frac{13}{4}}$

$x = \frac{5}{2} \pm \frac{\sqrt{13}}{\sqrt{4}}$

$= \frac{1}{2} \left(5 \pm \sqrt{13}\right)$

$x \approx 4.30278 \mathmr{and} \approx 0.69722$