How do you solve by completing the square: $x^{2} + 6 x - 5 = 0$ $x^2 + 6x - 5 = 0$ ?

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How do you solve by completing the square: $x^{2} + 6 x - 5 = 0$ $x^2 + 6x - 5 = 0$ ?

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Answer 1 · 2018-03-04T14:29:32

$x = - 3 + \sqrt{14} or x = - 3 - \sqrt{14}$ $x=-3+sqrt14orx=-3-sqrt14$

Explanation:

$x^{2} + 6 x - 5 = 0$ $x^2+6x-5=0$
$\Rightarrow x^{2} + 6 x + 9 - 14 = 0$ $rArrx^2+6x+color(red)(9-14)=0$
$\Rightarrow (x^{2} + 6 x + 9) = 14$ $rArr(x^2+6x+9)=14$
$\Rightarrow {(x + 3)}^{2} = {(\sqrt{14})}^{2}$ $rArrcolor(blue)((x+3)^2=(sqrt14)^2)$
$\Rightarrow x + 3 = \pm \sqrt{14}$ $rArrx+3=+-sqrt14$
$\Rightarrow x + 3 = \sqrt{14} or x + 3 = - \sqrt{14}$ $rArrx+3=sqrt14orx+3=-sqrt14$
$\Rightarrow x = - 3 + \sqrt{14} or x = - 3 - \sqrt{14}$ $rArrx=-3+sqrt14orx=-3-sqrt14$

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How do you solve by completing the square: $x^{2} + 6 x - 5 = 0$ $x^2 + 6x - 5 = 0$ ?

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Explanation:

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How do you solve by completing the square: $x^{2} + 6 x - 5 = 0$ $x^2 + 6x - 5 = 0$ ?