Asked by htht
Which of the following is an extraneous solution of StartRoot negative 3 x minus 2 EndRoot = x + 2?
x = –6
x = –1
x = 1
x = 6
All Answers 1
Answered by
GPT-5 mini
AI
Solve sqrt(-3x - 2) = x + 2. Domain: -3x - 2 >= 0 and x + 2 >= 0, so -2 <= x <= -2/3.
Square both sides: -3x - 2 = (x+2)^2 => x^2 +7x +6 = 0 => (x+1)(x+6)=0, so x = -1 or x = -6.
Check in the original: x = -1 works (sqrt(1)=1 = -1+2). x = -6 does not (sqrt(16)=4 ≠ -6+2 = -4), and x = -6 is outside the domain.
Thus the extraneous solution is x = -6.
Square both sides: -3x - 2 = (x+2)^2 => x^2 +7x +6 = 0 => (x+1)(x+6)=0, so x = -1 or x = -6.
Check in the original: x = -1 works (sqrt(1)=1 = -1+2). x = -6 does not (sqrt(16)=4 ≠ -6+2 = -4), and x = -6 is outside the domain.
Thus the extraneous solution is x = -6.
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