How do you solve using the completing the square method $3 x^{2} - 5 x - 6 = 0$ $3x^2-5x-6=0$ ?

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How do you solve using the completing the square method $3 x^{2} - 5 x - 6 = 0$ $3x^2-5x-6=0$ ?

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Answer 1 · 2016-05-10T20:25:30

$x = \frac{5}{6} \pm \frac{\sqrt{97}}{6}$ $x = 5/6+-sqrt(97)/6$

Explanation:

Use the difference of squares identity:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2=(a-b)(a+b)$

with $a = (6 x - 5)$ $a=(6x-5)$ and $b = \sqrt{97}$ $b=sqrt(97)$ .

To avoid too much arithmetic with fractions, pre-multiply the equation by $3 \cdot 2^{2} = 12$ $3*2^2 = 12$ first:

$0 = 12 (3 x^{2} - 5 x - 6)$ $0 = 12(3x^2-5x-6)$

$= 36 x^{2} - 60 x - 72$ $=36x^2-60x-72$

$= {(6 x - 5)}^{2} - 5^{2} - 72$ $=(6x-5)^2-5^2-72$

$= {(6 x - 5)}^{2} - 25 - 72$ $=(6x-5)^2-25-72$

$= {(6 x - 5)}^{2} - 97$ $=(6x-5)^2-97$

$= {(6 x - 5)}^{2} - {(\sqrt{97})}^{2}$ $=(6x-5)^2-(sqrt(97))^2$

$= ((6 x - 5) - \sqrt{97}) ((6 x - 5) + \sqrt{97})$ $=((6x-5)-sqrt(97))((6x-5)+sqrt(97))$

$= (6 x - 5 - \sqrt{97}) (6 x - 5 + \sqrt{97})$ $=(6x-5-sqrt(97))(6x-5+sqrt(97))$

$= 36 (x - \frac{5}{6} - \frac{\sqrt{97}}{6}) (x - \frac{5}{6} + \frac{\sqrt{97}}{6})$ $=36(x-5/6-sqrt(97)/6)(x-5/6+sqrt(97)/6)$

So $x = \frac{5}{6} \pm \frac{\sqrt{97}}{6}$ $x = 5/6+-sqrt(97)/6$

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How do you solve using the completing the square method $3 x^{2} - 5 x - 6 = 0$ $3x^2-5x-6=0$ ?

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How do you solve using the completing the square method $3 x^{2} - 5 x - 6 = 0$ $3x^2-5x-6=0$ ?