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How to solve for x in x² + 4x + 4 = 8 Explanation will get brainliest. Thanks guys!


x=−2±2√2

​ x=−2±4√ ​2

​ x=−4±2√ ​2

​ x=−4±4√ ​2

Respuesta :

wegnerkolmp2741o

Answer:

x = -2 ±2sqrt(2)

Step-by-step explanation:

x² + 4x + 4 = 8

Lets solve by completing the square

The left is set for completing the square since (4/2)^2 =4  so we rewrite this as (x+4/2)^2

(x+2) ^2 = 8

Take the square root of each side

sqrt((x+2) ^2 ) = ±sqrt(8)

x+2 =±sqrt(4*2)

x+2  =±sqrt(4)sqrt(2)

x+2 = ±2sqrt(2)

Subtract 2 from each side

x+2-2 = -2 ±2sqrt(2)

x = -2 ±2sqrt(2)

Wolfyy

Steps to solve:

x² + 4x + 4 = 8

~Subtract 8 to both sides

x² + 4x + 4 - 8 = 8 - 8

~Simplify

x² + 4x - 4 = 0

This is in standard form [ ax² + bx + c ], so we can find the a, b,  and c values.

a = 1

b = 4

c = -4

Use the quadratic formula [ [tex]x=\frac{-b+-\sqrt{b^2+4ac} }{2a}[/tex] ] to solve.

[tex]x = \frac{-4+-\sqrt{(4)^2-4(1)(-4)} }{2(1)}[/tex]

[tex]x = \frac{-4+-\sqrt{32} }{2}[/tex]

[tex]x = -2+-2\sqrt{2}[/tex]

Therefore, the correct answer is x = −2 ± 2√2 (Option A)

Best of Luck!