Eloise started to solve a radical equation in this way: Square root of negative 2x plus 1 − 3 = x Square root of negative 2x plu
s 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?
2 answers:
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
<span> the solution is x=-1.172 </span>
see attached graph to check
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