+0  
 
0
457
3
avatar+7485 

I need a solution.

 

x^2−8=√4x^2 laugh

asinus  Jul 21, 2017
edited by asinus  Jul 21, 2017
 #1
avatar+17745 
+2

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.

geno3141  Jul 21, 2017
 #2
avatar+7485 
+1

Thanks geno3141! That helped me.laugh

asinus  Jul 21, 2017
 #3
avatar+89803 
+4

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

 

 

cool cool cool

CPhill  Jul 22, 2017

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