How do you solve and check for extraneous solutions in #sqrt(x+1) + 5 = x#?

Question

Nathan Axel · Accepted Answer

The only valid solution is #x=8#
A second candidate solution #x=3# can be eliminated by checking for the validity of the given equation with #x=3# (and noting that it is not valid).

Given #sqrt(x+1) + 5 = x# Subtract #5# from both sides #color(white)("XXXX")##sqrt(x+1) = x-5# Square both sides (possibly generating an extraneous root at this point) #color(white)("XXXX")##x+1 = x^2-10x+25# Subtract #(x+1)# from both sides (and flip sides) #color(white)("XXXX")##x^2-11x+24 = 0# Factor #color(white)("XXXX")##(x-3)(x-8) = 0# #rArr##color(white)("XXXX")##x=3##color(white)("XXXX")#or#color(white)("XXXX")##x=8# Substituting #3# for #x# in the Left Side of original equation #color(white)("XXXX")##sqrt(3+1)+5 = 2+3# #color(white)("XXXX")##color(white)("XXXX")##!=3# The solution #x=3# is extraneous. Substituting #8# for #x# in the Left Side of original equation #color(white)("XXXX")##sqrt(8+1)+5 = 3+5# #color(white)("XXXX")##color(white)("XXXX")##=8# #color(white)("XXXX")##color(white)("XXXX")##=x# The solution #x=8# is valid.

Ethan Adkins · Answer

undefined To solve and check for extraneous solutions in the equation sqrt(x+1) + 5 = x, follow these steps:

1. Start by isolating the square root term on one side of the equation. Subtract 5 from both sides: sqrt(x+1) = x - 5.

2. Square both sides of the equation to eliminate the square root: (sqrt(x+1))^2 = (x - 5)^2. This simplifies to x + 1 = x^2 - 10x + 25.

3. Rearrange the equation to form a quadratic equation: x^2 - 11x + 24 = 0.

4. Factor the quadratic equation: (x - 3)(x - 8) = 0.

5. Set each factor equal to zero and solve for x: x - 3 = 0 or x - 8 = 0. This gives x = 3 or x = 8.

6. Check each solution by substituting them back into the original equation. For x = 3: sqrt(3+1) + 5 = 3, which simplifies to 4 + 5 = 3, which is false. For x = 8: sqrt(8+1) + 5 = 8, which simplifies to 6 + 5 = 8, which is also false.

7. Since both solutions result in false statements, there are no valid solutions to the equation. Therefore, there are no extraneous solutions to check for.