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1+[square root] n+11=n

A, -2 and 5 are both real solutions
B. -2 is a real solution and 5 is an extraneous solution
C. -2 is an extraneous solution and 5 is a real solution
D. -2 and 5 are both extraneous solutions

User Dfr
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1 Answer

4 votes

Answer:

To solve the equation, we need to isolate the square root term and square both sides of the equation. We have:

1 + √n + 11 = n

√n = n - 12

Squaring both sides:

n = (n - 12)^2

n = n^2 - 24n + 144

n^2 - 25n + 144 = 0

Factorizing the quadratic equation:

(n - 16)(n - 9) = 0

Therefore, n = 16 or n = 9. We need to check these solutions in the original equation to see if they are valid.

For n = 16:

1 + √16 + 11 = 16

1 + 4 + 11 = 16

16 = 16

The solution n = 16 satisfies the equation.

For n = 9:

1 + √9 + 11 = 9

1 + 3 + 11 = 9

15 = 9

The solution n = 9 does not satisfy the equation.

Therefore, the only real solution is n = 16, which corresponds to option C: -2 is an extraneous solution and 5 is a real solution.

Hope This Helps!

User David Kasabji
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