How do you solve $x^{2} + 3 x + 6 = 0$ $x^2 + 3x +6 =0$ by completing the square?

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How do you solve $x^{2} + 3 x + 6 = 0$ $x^2 + 3x +6 =0$ by completing the square?

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Answer 1 · 2016-01-28T13:48:18

$x = \pm \sqrt{3} - 3$ $x = ± sqrt3 - 3$

Explanation:

By adding 9 to both sides of the equation to obtain:

$(x^{2} + 3 x + 9) + 6 = 9$ $( x^2 + 3x + 9 )+ 6 = 9$

now ${(x + 3)}^{2} = 9 - 6 = 3$ $(x + 3 )^ 2 = 9-6 = 3$

${(x + 3)}^{2} is a perfect square$ $( x + 3 )^2 color(black)( " is a perfect square ")$

Taking the 'square root' of both sides :

$\sqrt{{(x + 3)}^{2}} = \sqrt{3}$ $sqrt((x+ 3 )^2 )= sqrt3$

hence x + 3 = ± $\sqrt{3}$ $sqrt3$

so x = ± $\sqrt{3} - 3$ $sqrt3 - 3$

Answer 2 · 2016-01-28T13:53:27

there are two $x$ $x$ values

$x_{1} = \frac{- 3 + \sqrt{15} i}{2}$ $x_1=(-3+sqrt(15)i)/2$

$x_{2} = \frac{- 3 - \sqrt{15} i}{2}$ $x_2=(-3-sqrt(15)i)/2$

Explanation:

Completing the square method:
Do this only when the numerical coefficient of $x^{2}$ $x^2$ is $1$ $1$ .
Start with the numerical coefficient of $x$ $x$ which is the number $3$ $3$ .
Divide this number by 2 then square the result. That is

${(\frac{3}{2})}^{2} = \frac{9}{4}$ $(3/2)^2=9/4$

Add $\frac{9}{4}$ $9/4$ to both sides of the equation

$x^{2} + 3 x + \frac{9}{4} + 6 = 0 + \frac{9}{4}$ $x^2+3x+9/4+6=0+9/4$

the first three terms now become one group which is a PST-Perfect Square Trinomial

$(x^{2} + 3 x + \frac{9}{4}) + 6 = \frac{9}{4}$ $(x^2+3x+9/4)+6=9/4$

${(x + \frac{3}{2})}^{2} + 6 = \frac{9}{4}$ $(x+3/2)^2+6=9/4$

${(x + \frac{3}{2})}^{2} = \frac{9}{4} - 6$ $(x+3/2)^2=9/4-6$ after transposing the $6$ $6$ to the right side

${(x + \frac{3}{2})}^{2} = \frac{9 - 24}{4}$ $(x+3/2)^2=(9-24)/4$

$\sqrt{{(x + \frac{3}{2})}^{2}} = \pm \sqrt{\frac{9 - 24}{4}}$ $sqrt((x+3/2)^2)=+-sqrt((9-24)/4)$

$x + \frac{3}{2} = \pm \sqrt{\frac{- 15}{4}}$ $x+3/2=+-sqrt((-15)/4)$

$x + \frac{3}{2} = \pm \frac{\sqrt{- 15}}{\sqrt{4}}$ $x+3/2=+-sqrt(-15)/sqrt(4)$

$x + \frac{3}{2} = \pm \frac{\sqrt{- 15}}{2}$ $x+3/2=+-sqrt(-15)/2$

Finally, transpose the $\frac{3}{2}$ $3/2$ to the right side of the equation

$x = - \frac{3}{2} \pm \frac{\sqrt{- 15}}{2}$ $x=-3/2+-sqrt(-15)/2$

take note: $\sqrt{- 15} = \sqrt{15} \cdot \sqrt{- 1} = \sqrt{15} i$ $sqrt(-15)=sqrt(15)*sqrt(-1)=sqrt(15)i$

therefore

$x = - \frac{3}{2} \pm \frac{\sqrt{15} i}{2}$ $x=-3/2+-(sqrt(15)i)/2$

there are two $x$ $x$ values

$x_{1} = \frac{- 3 + \sqrt{15} i}{2}$ $x_1=(-3+sqrt(15)i)/2$

$x_{2} = \frac{- 3 - \sqrt{15} i}{2}$ $x_2=(-3-sqrt(15)i)/2$

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How do you solve $x^{2} + 3 x + 6 = 0$ $x^2 + 3x +6 =0$ by completing the square?

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