4^2= (x+2)^2 solve for x

For x + 2 squared to equal 4 squared, x + 2 can equal either 4 or -4:

x + 2 = 4 ==> x = 2

x + 2 = -4 ==> x = -6


Check:

(2 + 2)^2 = 4^2 = 16

(-6 + 2)^2 = (-4)^2 = 16

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Here are 3 ways to approach this:

➊ 4² = (x+2)² ... Notice that both 4 and (x+2) are squared and that (x+2) can be
                          either positive or negative so that 4² = (x+2)² implies that 4 = ±(x+2).

                                                          4 = ±(x+2)
                                            4 = +(x+2)   OR   4 = -(x+2)
                                            2 = x                  4 = -x -2
                                            x = 2          OR   x = -6

                                                       ANSWER
                                                         x = -6, 2



➋        4² = (x+2)² ... Take the square root of both sides.
       √(4²) = √( (x+2)² )            ← √(u²) = │u│for all real numbers u
           4 = │x+2│
          ±4 = x+2                  ← If │u│=a, then u = ±a
     -2 ± 4 = x
      2, -6 = x                    ← ANSWER


➌ 4² = (x+2)² ... Expand
   16 = x² + 4x + 4
     0 = x² + 4x - 12
     0 = (x+6)(x-2)
          x= -6, x=2              ← ANSWER



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4^2 = (x + 2)^2
sqrt(4^2) = sqrt[(x + 2)^2]
+/- 4 = x + 2
x = 4 - 2 and x = - 4 - 2
x = 2 and x = - 6
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OR

4^2 = (x + 2)^2
16 = 4^2 = (x + 2)^2 = x^2 + 4x + 4
x^2 + 4x - 12 = 0
(x + 6).(x - 2) = 0
x + 6 = 0 or x - 2 = 0
x = - 6 or x = 2
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Hope this helps!

Since both sides are raised to the 2 power, you can just cancel those out... then 4 = (x+2) and x=2

16 = (x+2)^2

Square root both sides:

4 = x + 2
x = 2

x + 2 = +/- 4 => x = 2, -6