Respuesta :

carlosego

For this case we must solve the following equation by completing squares:

[tex]x ^ 2 + 6x-6 = 0[/tex]

We add 6 to both sides of the equation:

[tex]x ^ 2 + 6x = 6[/tex]

We divide the middle term by 2, and square it:

[tex](\frac {6} {2}) ^ 2[/tex]

And we add it to both sides of the equation:

[tex]x ^ 2 + 6x + (\frac {6} {2}) ^ 2 = 6 + (\frac {6} {2}) ^ 2\\x ^ 2 + 6x + (3) ^ 2 = 6 + 9[/tex]

We rewrite the left part of the equation:

[tex](x + 3) ^ 2 = 15[/tex]

We apply root to both sides:

[tex]x + 3 = \pm \sqrt {15}[/tex]

We have two solutions:

[tex]x_ {1} = \sqrt {15} -3\\x_ {2} = - \sqrt {15} -3[/tex]

Answer:

[tex]x_ {1} = \sqrt {15} -3\\x_ {2} = - \sqrt {15} -3[/tex]

luisejr77

Answer:

[tex]x=-3\±\sqrt{15}[/tex]

Step-by-step explanation:

We have the following equation

[tex]x^2+6x-6=0[/tex]

To use the method of completing squares you must take the coefficient of x and divide it by 2 and square the result.

[tex](\frac{6}{2})^2=9[/tex]

Now add 9 on both sides of equality

[tex](x^2+6x+ 9)-6=9[/tex]

Factor the term in parentheses

[tex](x+3)^2-6=9[/tex]

Add 6 on both sides of the equation

[tex](x+3)^2-6+6=9+6[/tex]

[tex](x+3)^2=15[/tex]

Take square root on both sides of the equation

[tex]\sqrt{(x+3)^2}=\±\sqrt{15}[/tex]

[tex]x+3=\±\sqrt{15}[/tex]

Subtract 3 from both sides of the equation.

[tex]x+3-3=-3\±\sqrt{15}[/tex]

[tex]x=-3\±\sqrt{15}[/tex]

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