How do you solve the equation and identify any extraneous solutions for #sqrt(x+7) = x + 1#?

Question

Avery Alexander · Accepted Answer

The term under the root must be #>=0# so #x>=-7#
The outcome must also be #>0# so #x>=-1# Now we can square both sides: #x+7=(x+1)^2=x^2+2x+1-># #x^2+2x-x+1-7=x^2+x-6=0# #(x+3)(x-2)=0->x=-3orx=2# The only allowed solution is #x=2# as the other one does not meet the conditions at the start.

Nathan Axel · Answer

undefined To solve the equation sqrt(x+7) = x + 1 and identify any extraneous solutions, we can follow these steps:

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

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

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

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

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

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

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

8. 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.

9. 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.