How do you solve using the completing the square method $x^2 + 3x – 6 = 0$ ?

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How do you solve using the completing the square method $x^2 + 3x – 6 = 0$ ?

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Answer 1 · 2018-06-26T10:00:48

$x=-3/2+-sqrt33/2$

Explanation:

$x^2+3x=6$

$x^2+2(3/2)xcolor(red)(+9/4)=6color(red)(+9/4)$

$(x+3/2)^2=33/4$

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

$sqrt((x+3/2)^2)=+-sqrt(33/4)larrcolor(blue)"note plus or minus"$

$x+3/2=+-sqrt33/2$

$"subtract "3/2" from both sides"$

$x=-3/2+-sqrt33/2larrcolor(red)"exact solutions"$

Answer 2 · 2018-06-26T10:01:27

$color(brown)(x = (-1/2) (3 - sqrt33 ), -(1/2) ( 3 + sqrt33)$

Explanation:

$x^2 + 3x - 6 = 0$

$x^2 + 3x + (3/2)^2 - 6 = (3/2)^2, "adding " (3/2)^2 " to both sides"$

$x^2 + 2 * 3/2 * x + (3/2)^2 = (3/2)^2 + 6$

$(x + 3/2)^2 = 33/4$

$(x + 3/2)^2 = (sqrt(33/4))^2$

$x + 3/2 = +- sqrt(33/4), " taking squre root on both sides"$

$x = -3/2 +- sqrt33 / 2$

$color(brown)(x = (-1/2) (3 - sqrt33 ), -(1/2) ( 3 + sqrt33)$

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How do you solve using the completing the square method $x^2 + 3x – 6 = 0$ ?

2 Answers

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How do you solve using the completing the square method x^2 + 3x – 6 = 0x^2 + 3x – 6 = 0?

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How do you solve using the completing the square method $x^2 + 3x – 6 = 0$ ?