How do you solve using the completing the square method $x^{2} + 10 x - 2 = 0$ $x^2+10x-2=0$ ?

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Answer 1 · 2016-05-28T17:06:53

The solutions are:
$x = 3 \sqrt{3} - 5$ $color(green)(x = 3sqrt 3 - 5$ , $x = - 3 \sqrt{3} - 5$ $color(green)(x = -3sqrt 3 -5$

$x^{2} + 10 x - 2 = 0$ $x^2 +10x - 2 = 0$

$x^{2} + 10 x = 2$ $x^2 +10x = 2$

To write the Left Hand Side as a Perfect Square, we add 25 to both sides:

$x^{2} + 10 x + 25 = 2 + 25$ $x^2 +10x + color(blue)(25)= 2 + color(blue)(25)$

$x^{2} + 2 \cdot x \cdot 5 + 5^{2} = 27$ $x^2 + 2 * x * 5 + 5^2 = 27$

Using the Identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $color(blue)((a+b)^2 = a^2 + 2ab + b^2$ , we get

${(x + 5)}^{2} = 27$ $(x+5)^2 = 27$

$x + 5 = \sqrt{27}$ $x + 5 = sqrt27$ or $x + 5 = - \sqrt{27}$ $x +5 = -sqrt27$

(Note: prime factorising #27; 27 = 3 * 3 * 3 = 3^3

So, $\sqrt{27} = \sqrt{3^{3}} = 3 \sqrt{3}$ $sqrt27 = sqrt (3^3) = 3sqrt3$ )

$x + 5 = 3 \sqrt{3}$ $x + 5 = 3sqrt3$ or $x + 5 = - 3 \sqrt{3}$ $x +5 = -3sqrt3$

$x = 3 \sqrt{3} - 5$ $color(green)(x = 3sqrt 3 - 5$ , $x = - 3 \sqrt{3} - 5$ $color(green)(x = -3sqrt 3 -5$

How do you solve using the completing the square method x^2+10x-2=0x2+10x−2=0x^2+10x-2=0?