How do you solve $3 x^{2} + 14 x = 5$ $3x^2+14x=5$ by completing the square?

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Answer 1 · 2015-05-12T19:08:16

Dividing through by $3$ $3$ , we get:

$x^{2} + (\frac{14}{3}) x = \frac{5}{3}$ $x^2+(14/3)x = 5/3$

Now try squaring $x + \frac{7}{3}$ $x+7/3$

${(x + \frac{7}{3})}^{2} = (x + \frac{7}{3}) (x + \frac{7}{3})$ $(x+7/3)^2 = (x+7/3)(x+7/3)$

$= x^{2} + (\frac{14}{3}) x + \frac{49}{9}$ $= x^2 + (14/3)x + 49/9$

$= \frac{5}{3} + \frac{49}{9}$ $= 5/3 + 49/9$

$= \frac{15}{9} + \frac{49}{9}$ $= 15/9+49/9$

$= \frac{64}{9}$ $= 64/9$

$= \frac{8^{2}}{3^{2}}$ $= (8^2)/(3^2)$

Hence $x + \frac{7}{3} = \pm \frac{8}{3}$ $x + 7/3 = +-8/3$

Subtracting 7/3 from both sides, we get

$x = \pm \frac{8}{3} - \frac{7}{3}$ $x = +-8/3 - 7/3$

So $x = \frac{1}{3}$ $x = 1/3$ or $x = - \frac{15}{3} = - 5$ $x = -15/3 = -5$ .

How do you solve 3x2+14x=53x^2+14x=5 by completing the square?