How do you solve $x^{2} + 8 x + 2 = 0$ $x^2 + 8x + 2 = 0$ by completing the square?

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How do you solve $x^{2} + 8 x + 2 = 0$ $x^2 + 8x + 2 = 0$ by completing the square?

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Answer 1 · 2016-04-29T16:58:32

$x = - 4 \pm \sqrt{14}$ $x = -4+-sqrt(14)$

Explanation:

We will use the difference of squares identity, which can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

with $a = (x + 4)$ $a=(x+4)$ and $b = \sqrt{14}$ $b=sqrt(14)$ as follows:

$0 = x^{2} + 8 x + 2$ $0 = x^2+8x+2$

$= {(x + 4)}^{2} - 16 + 2$ $=(x+4)^2-16+2$

$= {(x + 4)}^{2} - {(\sqrt{14})}^{2}$ $=(x+4)^2-(sqrt(14))^2$

$= ((x + 4) - \sqrt{14}) ((x + 4) + \sqrt{14})$ $=((x+4)-sqrt(14))((x+4)+sqrt(14))$

$= (x + 4 - \sqrt{14}) ((x + 4 + \sqrt{14})$ $=(x+4-sqrt(14))((x+4+sqrt(14))$

Hence:

$x = - 4 \pm \sqrt{14}$ $x = -4+-sqrt(14)$

Answer 2 · 2016-05-02T00:49:19

$x = \sqrt{14} - 4$ $x = sqrt14 -4$ OR $x = - \sqrt{14} - 4$ $x = -sqrt14 -4$
$x = - 0.258$ $x = -0.258$ OR $x = - 7.742$ $x = -7.742$ (3 dec places)

Explanation:

Completing the square is based on the consistency of the answers to the square of a binomial.

${(x - 3)}^{2} = x^{2} - 6 x + 9$ $(x - 3)^2 = x^2 - 6x + 9$
${(x - 5)}^{2} = x^{2} - 10 x + 25$ $(x - 5)^2 = x^2 - 10x + 25$
${(x + 6)}^{2} = x^{2} + 12 x + 36$ $(x + 6)^2 = x^2 + 12x + 36$
In all of the products above, $a x^{2} + b x + c$ $ax^2+ bx + c$ we see the following:

$a = 1$ $a = 1$
The first and last terms, $a and c$ $a and c$ are perfect squares.'
There is a specific relationship between 'b' - the coefficient of the $x$ $x$ term and 'c'. Half of b, squared equals c.

Knowing this, it is always possible to add in a missing value for $c$ $c$ to have the square of a binomial, which can then be written as ${(x + ?)}^{2}$ $(x+?)^2$

In $x^{2} + 8 x + 2 = 0$ $x^2 + 8x + 2 = 0$ , 2 is obviously not the correct value of $c$ $c$ .
It is therefore moved to the right hand side and the wanted value of $c$ $c$ is added to BOTH sides of the equation.

$x^{2} + 8 x + 16$ $x^2 + 8x + color(red)(16)$ = $- 2 + 16 \Rightarrow [16 = {(8 \div 2)}^{2}]$ $-2 + color(red)(16) rArr [16 = (8÷2)^2]$
${(x + 4)}^{2} = 14 \Rightarrow$ $(x + 4)^2 = 14 rArr$ where 4 is either from $b \div 2 or \sqrt{16}$ $b÷2 or sqrt16$
$x + 4 = \pm \sqrt{14} \Rightarrow$ $x + 4 = +-sqrt14 rArr$ take the square root of both sides

This gives 2 possible answers for $x$ $x$ .

$x = \sqrt{14} - 4$ $x = sqrt14 -4$ OR $x = - \sqrt{14} - 4$ $x = -sqrt14 -4$

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How do you solve $x^{2} + 8 x + 2 = 0$ $x^2 + 8x + 2 = 0$ by completing the square?

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How do you solve $x^{2} + 8 x + 2 = 0$ $x^2 + 8x + 2 = 0$ by completing the square?