How do you solve the equation and identify any extraneous solutions for #sqrt(x+7) = x + 1#?
The term under the root must be
The outcome must also be
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To solve the equation sqrt(x+7) = x + 1 and identify any extraneous solutions, we can follow these steps:

Square both sides of the equation to eliminate the square root: (sqrt(x+7))^2 = (x + 1)^2.

Simplify the equation: x + 7 = x^2 + 2x + 1.

Rearrange the equation to form a quadratic equation: x^2 + x  6 = 0.

Factor the quadratic equation: (x + 3)(x  2) = 0.

Set each factor equal to zero and solve for x: x + 3 = 0 or x  2 = 0.

Solve for x in each equation: x = 3 or x = 2.

Check for extraneous solutions by substituting each potential solution back into the original equation.

For x = 3: sqrt((3) + 7) = 3 + 1. Simplifying, we get sqrt(4) = 2, which is not true. Therefore, x = 3 is an extraneous solution.

For x = 2: sqrt((2) + 7) = 2 + 1. Simplifying, we get sqrt(9) = 3, which is true. Therefore, x = 2 is the valid solution.
Therefore, the solution to the equation sqrt(x+7) = x + 1 is x = 2, and the extraneous solution is x = 3.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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