Asked by sharl
Solve the equation. Identify any extraneous solutions.
x = sqrt root 3x + 40
x = sqrt root 3x + 40
Answers
Answered by
Henry
X = sqrt(3X+40),
Square both sides:
X^2 = 3X + 40,
X^2 - 3X - 40 = 0,
X-8)(X+5) = 0,
X-5=0,
X = 5.
X+5 = 0,
X = -5.
Solution set: X = -5, and X = 5.
5 does not satisfy the original Eq.
Therefore, it is an extraneous solution.
Square both sides:
X^2 = 3X + 40,
X^2 - 3X - 40 = 0,
X-8)(X+5) = 0,
X-5=0,
X = 5.
X+5 = 0,
X = -5.
Solution set: X = -5, and X = 5.
5 does not satisfy the original Eq.
Therefore, it is an extraneous solution.
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