How do you solve $2 x^{2} + 12 x - 14 = 0$ $2x^2 + 12x − 14 = 0$ by completing the square?

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How do you solve $2 x^{2} + 12 x - 14 = 0$ $2x^2 + 12x − 14 = 0$ by completing the square?

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Answer 1 · 2017-02-12T14:18:04

$x = - 7 XX or XX x = + 1$ $x=-7color(white)("XX")orcolor(white)("XX")x=+1$

Explanation:

$2 x^{2} + 12 x - 14 = 0$ $2x^2+12x-14=0$

$\to 2 x^{2} + 12 x = 14$ $rarr2x^2+12x=14$

$\to x^{2} + 6 x = 7$ $rarr x^2+6x=7$

Now to "complete the square":
If $x^{2} + 6 x$ $x^2+6x$ are the first two terms of the expansion of a squared binomial: ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2 = x^2+2ax+a^2$
then $2 a x$ $2ax$ must equal $6 x$ $6x$ ;
that is $a = 3$ $a=3$ and $a^{2} = 9$ $a^2=9$

To "complete the square" we must add $9$ $color(magenta)(9)$ to the expression,
but we can only legally do this if we add $9$ $color(magenta)9$ to both sides of the equation:
$XXX x^{2} + 6 x + 9 = 7 + 9$ $color(white)("XXX")x^2+6xcolor(magenta)(+9) = 7color(magenta)(+9)$

$XXX \to {(x + 3)}^{2} = 16$ $color(white)("XXX")rarr (x+3)^2=16$

$XXX \to x + 3 = \pm 4$ $color(white)("XXX")rarr x+3=+-4$

$XXX \to x = - 3 \pm 4$ $color(white)("XXX")rarr x=-3+-4$

$XXX$ $color(white)("XXX")$ which can be written as: $x = - 7 X or X x = 1$ $x=-7color(white)("X")orcolor(white)("X")x=1$

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How do you solve $2 x^{2} + 12 x - 14 = 0$ $2x^2 + 12x − 14 = 0$ by completing the square?

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Explanation:

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How do you solve 2x^2 + 12x − 14 = 02x2+12x−14=0 2x^2 + 12x − 14 = 0 by completing the square?

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How do you solve $2 x^{2} + 12 x - 14 = 0$ $2x^2 + 12x − 14 = 0$ by completing the square?