Solve the equation. Identify any extraneous solutions.
x = sqrt root 3x + 40
To solve the equation x = √(3x + 40), we can follow these steps:
Step 1: Square both sides of the equation to eliminate the square root:
x^2 = (3x + 40)
Step 2: Expand the equation:
x^2 = 3x + 40
Step 3: Rearrange the equation into a standard quadratic form:
x^2 - 3x - 40 = 0
Step 4: Factor the quadratic equation:
(x - 8)(x + 5) = 0
Step 5: Set each factor equal to zero and solve for x:
x - 8 = 0 or x + 5 = 0
Solving x - 8 = 0 gives x = 8, and solving x + 5 = 0 gives x = -5.
So the two possible solutions for x are 8 and -5.
However, we need to check if either of these solutions lead to any extraneous solutions. An extraneous solution is when the solution found does not satisfy the original equation.
Plugging x = 8 back into the original equation:
√(3(8) + 40) = √(24 + 40) = √(64) = 8
The solution x = 8 is valid.
Plugging x = -5 back into the original equation:
√(3(-5) + 40) = √(-15 + 40) = √(25) = 5
The solution x = -5 is also valid.
Thus, both x = 8 and x = -5 are valid solutions to the equation, and there are no extraneous solutions in this case.
To solve the equation x = sqrt(root(3x + 40)), we need to follow these steps:
Step 1: Start by squaring both sides of the equation to eliminate the square root:
(x)^2 = (sqrt(root(3x + 40)))^2
x^2 = root(3x + 40)
Step 2: Square both sides again to eliminate the root:
(x^2)^2 = (root(3x + 40))^2
x^4 = 3x + 40
Step 3: Rearrange the equation so that one side equals zero:
x^4 - 3x - 40 = 0
Step 4: Factorize the quadratic equation:
(x^2 + 5)(x^2 - 8) = 0
Step 5: Set each factor equal to zero and solve for x:
x^2 + 5 = 0
x^2 = -5
x = ±√(-5) (This solution is imaginary, so it is extraneous.)
x^2 - 8 = 0
x^2 = 8
x = ±√8
x = ±2√2
Therefore, the solutions to the equation are x = ±2√2, with the extraneous solution x = ±√(-5).
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.