Question

Eloise started to solve a radical equation in this way: Square root of negative 2x plus 1 − 3 = x Square root of negative 2x plus 1 − 3 + 3 = x + 3 Square root of negative 2x plus 1 = x + 3 Square root of negative 2x plus 1 − 1 = x + 3 − 1 Square root of negative 2 x = x + 2 ( Square root of negative 2 x )2 = (x − 4)2 −2x = x2 − 8x + 16 −2x + 2x = x2 + 8x + 16 + 2x 0 = x2 + 10x + 16 0 = (x + 2)(x + 8) x + 2 = 0 x + 8 = 0 x + 2 − 2 = 0 − 2 x + 8 − 8 = 0 − 8 x = −2 x = −8 Both solutions are extraneous because they don't satisfy the original equation. What error did Eloise make?

Answer 1

 Square root of negative 2x plus 1 − 3 = x

 ((-2x+1 ^1/2)-3=x

Eloise procedure

((-2x+1) ^1/2)-3=x------ok

((-2x+1) ^1/2)-3+3=x+3-----ok

((-2x+1) ^1/2)=x+3-----------ok

((-2x+1-1) ^1/2)=x+3-1---------->this is the error

 the correct is ((-2x+1) ^1/2)-1=x+3-1

 I solve correctly

((-2x+1) ^1/2)=x+3------------((-2x+1) ^1/2) ^2=(x+3) ^2

(-2x+1) =(x+3) ^2

-2x+1=x^2+6x+9

 -x^2+8x+8=0

solving the quadratic equation

x1=-6.828

x2=-1.172

 the solution is x=-1.172 

see attached graph to check

Answer 2

Eloise made these steps well:
 Square root of negative 2x plus 1 - 3 = x
 Square root of negative 2x plus 1 - 3 + 3 = x + 3
 The error comes in the next step:
 Square root of negative 2x plus 1 = x + 3
 Square root of negative 2x plus 1 -1 = x + 3 - 1
 Square root of negative 2x  = x + 2
 the number one is inside the root, therefore they can not be subtracted.
 answer
 error in the step:
  Square root of negative 2x  = x + 2