How do you solve $x^{2} + 14 x - 15 = 0$ $x^2 + 14x - 15 = 0$ by completing the square?

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Answer 1 · 2016-04-04T16:30:00

The solutions are:
$x = 1$ $color(green)(x = 1$ or $x = - 15$ $color(green)(x = -15$

$x^{2} + 14 x - 15 = 0$ $x^2 + 14x - 15=0$

$x^{2} + 14 x = 15$ $x^2 + 14x = 15$

To write the Left Hand Side as a Perfect Square, we add $49$ $49$ to both sides:

$x^{2} + 14 x + 49 = 15 + 49$ $x^2 + 14x + 49 = 15 + 49$

$x^{2} + 2 \cdot x \cdot 7 + 7^{2} = 64$ $x^2 + 2* x * 7 + 7^2 = 64$

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

${(x + 7)}^{2} = 64$ $(x+7)^2 = 64$

$x + 7 = \sqrt{64}$ $x + 7 = sqrt64$ or $x + 7 = - \sqrt{64}$ $x +7 = -sqrt64$

$x = 8 - 7 = 1$ $color(green)(x = 8-7 = 1$ or $x = - 8 - 7 = - 15$ $color(green)(x = -8 - 7 = -15$

How do you solve x^2 + 14x - 15 = 0x2+14x−15=0 x^2 + 14x - 15 = 0 by completing the square?