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How do you solve $x^{2} + 10 x - 1 = 0$ $x^2+10x-1=0$ using completing the square?

1 Answer

Jun 15, 2015

$x = - 5 \pm \sqrt{26}$ $x = -5+-sqrt(26)$

Explanation:

$x^{2} + 10 x - 1 = 0$ $x^2+10x-1 = 0$

simplify by moving the constant to the right side
$XXXX$ $color(white)("XXXX")$ $x^{2} + 10 x = 1$ $x^2+10x = 1$

complete the square
$XXXX$ $color(white)("XXXX")$ $x^{2} + 10 x + {(\frac{10}{2})}^{2} = 1 + {(\frac{10}{2})}^{2}$ $x^2+10x+(10/2)^2 = 1 + (10/2)^2$

rewrite as a square and simplify the right side
$XXXX$ $color(white)("XXXX")$ ${(x + 5)}^{2} = 26$ $(x+5)^2 = 26$

take the square root of both sides
$XXXX$ $color(white)("XXXX")$ $x + 5 = \pm \sqrt{26}$ $x+5 = +-sqrt(26)$

subtract $5$ $5$ from both sides to isolate $x$ $x$
$XXXX$ $color(white)("XXXX")$ $x = - 5 \pm \sqrt{26}$ $x = -5+-sqrt(26)$

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