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

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

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Answer 1 · 2018-04-15T18:00:51

$x = - \frac{3}{2} \pm \frac{1}{2} \sqrt{19}$ $x=-3/2+-1/2sqrt19$

Explanation:

$using the method of completing the square$ $"using the method of "color(blue)"completing the square"$

$∙ the coefficient of the x^{2} term must be 1$ $• " the coefficient of the "x^2" term must be 1"$

$factor out 2$ $"factor out "2$

$\Rightarrow 2 (x^{2} + 3 x - \frac{5}{2}) = 0$ $rArr2(x^2+3x-5/2)=0$

$∙ add/subtract {(\frac{1}{2} coefficient of the x-term)}^{2} to$ $• " add/subtract "(1/2"coefficient of the x-term")^2" to"$
$x^{2} + 3 x$ $x^2+3x$

$\Rightarrow 2 (x^{2} + 2 (\frac{3}{2}) x + \frac{9}{4} - \frac{9}{4} - \frac{5}{2}) = 0$ $rArr2(x^2+2(3/2)xcolor(red)(+9/4)color(red)(-9/4)-5/2)=0$

$\Rightarrow 2 {(x + \frac{3}{2})}^{2} + 2 (- \frac{9}{4} - \frac{5}{2}) = 0$ $rArr2(x+3/2)^2+2(-9/4-5/2)=0$

$\Rightarrow 2 {(x + \frac{3}{2})}^{2} - \frac{19}{2} = 0$ $rArr2(x+3/2)^2-19/2=0$

$\Rightarrow 2 {(x + \frac{3}{2})}^{2} = \frac{19}{2}$ $rArr2(x+3/2)^2=19/2$

$\Rightarrow {(x + \frac{3}{2})}^{2} = \frac{19}{4}$ $rArr(x+3/2)^2=19/4$

$take the square root of both sides$ $color(blue)"take the square root of both sides"$

$\sqrt{{(x + \frac{3}{2})}^{2}} = \pm \sqrt{\frac{19}{4}} \leftarrow note plus or minus$ $sqrt((x+3/2)^2)=+-sqrt(19/4)larrcolor(blue)"note plus or minus"$

$\Rightarrow x + \frac{3}{2} = \pm \frac{1}{2} \sqrt{19}$ $rArrx+3/2=+-1/2sqrt19$

$\Rightarrow x = - \frac{3}{2} \pm \frac{1}{2} \sqrt{19} \leftarrow exact solutions$ $rArrx=-3/2+-1/2sqrt19larrcolor(red)"exact solutions"$

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

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Explanation:

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