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DiKsa [7]
3 years ago
13

Solve the quadratic equation by completing the square. 6x2 + 4x - 5 = 0

Mathematics
1 answer:
Nikitich [7]3 years ago
6 0

Answer:

$x_{1}=-\frac{2+\sqrt{34}}{6}$

$x_{2}=-\frac{2-\sqrt{34}}{6}$

Step-by-step explanation:

Quadratic Equation given:

6x^2 + 4x - 5 = 0

Start dividing both sides by 6.

$\frac{6}{6} x^2 +\frac{4}{6} x - \frac{5}{6}  = \frac{0}{6} $

$x^2 +\frac{2}{3} x - \frac{5}{6}  = 0 $

$x^2 +\frac{2}{3} x = \frac{5}{6} $

Once, $\left( \frac{2}{3} \cdot \frac{1}{2} \right)^2=\frac{1}{9} $

$x^2 +\frac{2}{3} x+\frac{1}{9}  = \frac{5}{6}+\frac{1}{9}  $

$\left(x+\frac{1}{3} \right)^2 = \frac{17}{18}}  $

$x+\frac{1}{3}  =\pm\sqrt{\frac{17}{18}} }   $

Solving $\sqrt{\frac{17}{18}} $

$\sqrt{\frac{17(18)}{18(18)}}= \sqrt{\frac{306}{324}}=\frac{3\sqrt{34} }{18} =\frac{\sqrt{34} }{6} $

Then,

$x+\frac{1}{3}  =\pm\frac{\sqrt{34} }{6}  $

$x =-\frac{1}{3} \pm\frac{\sqrt{34} }{6}  $

$x_{1}=-\frac{2+\sqrt{34}}{6}$

$x_{2}=-\frac{2-\sqrt{34}}{6}$

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