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

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

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Answer 1 · 2015-04-02T07:11:37

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

The general form of a trinomial is $a x^{2} + b x + c = 0$ $ax^2+bx+c=0"$ The letter $c$ $c$ is the constant.

Solve the trinomial $x^{2} + 8 x + 2 = 0$ $x^2+8x+2=0$

First move the constant to the right side by subtracting 2 from both sides.

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

Divide only the coefficient of 8x by 2. Square the result, and add that value to both sides of the equation.

${(\frac{8}{2})}^{2} = {(4)}^{2} = 16$ $((8)/(2))^2=(4)^2=16$

$x^{2} + 8 x + 16 = - 2 + 16$ $x^2+8x+16 = -2+16$

$x^{2} + 8 x + 16 = 14$ $x^2+8x+16=14$

The left side is now a perfect square trinomial. Factor the perfect square trinomial.

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

Take the square root of each side and solve.

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

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

Source:
http://www.regentsprep.org/regents/math/algtrig/ate12/completesqlesson.htm

Answer 2 · 2015-04-02T07:21:14

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

The general form of a trinomial is $a x^{2} + b x + c = 0$ $ax^2+bx+c=0"$ The letter $c$ $c$ is the constant.

Solve the trinomial $x^{2} + 8 x + 2 = 0$ $x^2+8x+2=0$

First move the constant to the right side by subtracting 2 from both sides.

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

Divide only the coefficient of 8x by 2. Square the result, and add that value to both sides of the equation.

${(\frac{8}{2})}^{2} = {(4)}^{2} = 16$ $((8)/(2))^2=(4)^2=16$

$x^{2} + 8 x + 16 = - 2 + 16$ $x^2+8x+16 = -2+16$

$x^{2} + 8 x + 16 = 14$ $x^2+8x+16=14$

The left side is now a perfect square trinomial. Factor the perfect square trinomial.

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

Take the square root of each side and solve.

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

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

Source:
http://www.regentsprep.org/regents/math/algtrig/ate12/completesqlesson.htm

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