Please help. Solution set.

Problem:  x² - 8  =  sqrt(4·x²)

     --->     x² - 8  =  sqrt(4) · sqrt(x²)

     --->     x² - 8  =  2· | x |

Split the absolute value into two parts:

Case 1:  If x >= 0, then | x | = x

     --->     x² - 8  =  2· | x |

     --->     x² - 8  =  2 · x

     --->     x² - 8  =  2x

     --->     x² - 2x - 8  =  0

     --->     (x - 4)(x + 2)  =  0

     --->     x = 4  or  x = -2

     --->     Since this case assumes that x >= 0, keep only the value  x = 4.

Case 2:  If x < 0, then | x | = -x

     --->     x² - 8  =  2· | x |

     --->     x² - 8  =  2 · -x

     --->     x² - 8  =  -2x

     --->     x² + 2x - 8  =  0

     --->     (x + 4)(x - 2)  =  0

     --->     x = -4  or  x = 2

     --->     Since this case assumes that x < 0, keep only the value  x = -4.

These are the two solutions.

x^2−8=√(4x^2)   square both sides

x^4 -16x^2 + 64  = 4x^2

X^4 - 20x^2 + 64 = 0     factor as

(x^2  - 16) ( x^2 - 4)  = 0

(x - 4) ( x + 4) (x - 2) ( x + 2)  = 0

Setting each factor to 0 and solving for x we have the possible solutions 

x = ± 2   and x = ± 4

Only the second pair actually solve the original equation....the other two are extraneous due to squaring both sides