How do you solve the quadratic equation by completing the square: $x^{2} - 12 x - 2 = 0$ $x^2 -12x-2=0$ ?

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How do you solve the quadratic equation by completing the square: $x^{2} - 12 x - 2 = 0$ $x^2 -12x-2=0$ ?

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Answer 1 · 2015-07-13T21:45:52

$x = 6 + \sqrt{38}$ $x= 6+sqrt(38)$ or $x = 6 - \sqrt{38}$ $x=6-sqrt(38)$
$XXXX$ $color(white)("XXXX")$ (using the completing the squares method)

Explanation:

Given $x^{2} - 12 x - 2 = 0$ $x^2-12x-2 = 0$

Completing the square is simpler if we move the constant to the right side:
$XXXX$ $color(white)("XXXX")$ $x^{2} - 12 x = 2$ $x^2-12x = 2$

If $x^{2}$ $x^2$ and $- 12 x$ $-12x$ are the first two terms of a squared binomial expansion:
$XXXX$ $color(white)("XXXX")$ ${(x - a)}^{2} = (x^{2} - 2 a x + a^{2})$ $(x-a)^2 = (x^2-2ax+a^2)$
then
$XXXX$ $color(white)("XXXX")$ $a = 6$ $a=6$
and
we need to add $a^{2} = 36$ $a^2 = 36$ to both sides to complete the square

$XXXX$ $color(white)("XXXX")$ $x^{2} - 12 x + 36 = 2 + 36$ $x^2-12x+36 = 2+36$

Rewriting as a squared binomial (and simplifying the right side)
$XXXX$ $color(white)("XXXX")$ ${(x - 6)}^{2} = 38$ $(x-6)^2 = 38$

Taking the square root of both sides
$XXXX$ $color(white)("XXXX")$ $x - 6 = \pm \sqrt{38}$ $x-6 = +-sqrt(38)$

Adding $6$ $6$ to both sides:
$XXXX$ $color(white)("XXXX")$ $x = 6 \pm \sqrt{38}$ $x=6+-sqrt(38)$

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How do you solve the quadratic equation by completing the square: $x^{2} - 12 x - 2 = 0$ $x^2 -12x-2=0$ ?

1 Answer

Explanation:

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How do you solve the quadratic equation by completing the square: x^2 -12x-2=0x2−12x−2=0x^2 -12x-2=0?

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Explanation:

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How do you solve the quadratic equation by completing the square: $x^{2} - 12 x - 2 = 0$ $x^2 -12x-2=0$ ?