How do you solve using the completing the square method $3 x^{2} - 7 x - 6 = 0$ $3x^2-7x-6=0$ ?

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How do you solve using the completing the square method $3 x^{2} - 7 x - 6 = 0$ $3x^2-7x-6=0$ ?

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Answer 1 · 2017-03-02T14:35:26

$x = 3$ $color(green)(x=3)$ or $x = - \frac{2}{3}$ $color(green)(x=-2/3)$
$XXX$ $color(white)("XXX")$ see below for solution method using "completing the square"

Explanation:

$3 x^{2} - 7 x - 6 = 0$ $3x^2-7x-6=0$
$\Rightarrow XXX x^{2} - \frac{7}{3} x = 2$ $rArrcolor(white)("XXX")color(blue)(x^2-7/3x)=color(blue)2$

If $x^{2} - \frac{7}{3} x$ $x^2-7/3x$ are the first two terms of a squared binomial:
$XXX {(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $color(white)("XXX")(x+a)^2=x^2+2ax+a^2$
then
$XXX 2 a = - \frac{7}{3} X \to X a = - \frac{7}{6}$ $color(white)("XXX")2a=-7/3color(white)("X")rarrcolor(white)("X")a=-7/6$
and to "complete the square" we will need to add
$XXX a^{2} = {(- \frac{7}{6})}^{2} = \frac{49}{36}$ $color(white)("XXX")color(red)(a^2)=(-7/6)^2=color(red)(49/36)$

giving:
$XXX x^{2} - \frac{7}{6} x + \frac{49}{36} = 2 + \frac{49}{36}$ $color(white)("XXX")color(blue)(x^2-7/6x)color(red)(+49/36)=color(blue)2color(red)(+49/36)$

$\Rightarrow XXX {(x - \frac{7}{6})}^{2} = \frac{2 \times 36 + 49}{36} = \frac{121}{36} = {(\frac{11}{6})}^{2}$ $rArrcolor(white)("XXX")(x-7/6)^2= (2xx36+49)/36=121/36= (11/6)^2$

$\Rightarrow XXX x - \frac{7}{6} = \pm \frac{11}{6}$ $rArrcolor(white)("XXX")x-7/6=+-11/6$

$\Rightarrow XXX x = \frac{7 \pm 11}{6}$ $rArrcolor(white)("XXX")x=(7+-11)/6$

$\Rightarrow XXX x = \frac{18}{6} = 3 X or X x = - \frac{4}{6} = - \frac{2}{3}$ $rArrcolor(white)("XXX")x=18/6=3color(white)("X")orcolor(white)("X")x=-4/6=-2/3$

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How do you solve using the completing the square method $3 x^{2} - 7 x - 6 = 0$ $3x^2-7x-6=0$ ?

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How do you solve using the completing the square method 3x2−7x−6=03x^2-7x-6=0?

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How do you solve using the completing the square method $3 x^{2} - 7 x - 6 = 0$ $3x^2-7x-6=0$ ?