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How many true, real number solutions does the equation n + 2 = √-16-5n have?
solution(s)

User Zamboney
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1 Answer

4 votes

Answer:

none

Explanation:

The domain of the equation can be found by looking at the requirements ...

  • the square root is non-negative
  • the argument of the square root is non-negative.

Domain

For n+2 ≥ 0, we find ...

n ≥ -2 . . . . . . . . subtract 2 from both sides

For -16-5n ≥ 0, we find ...

-16 ≥ 5n . . . . . add 5n

-3.2 ≥ n . . . . . divide by 5

Together, these domain restrictions require that ...

n ≥ -2

n ≤ -3.2

These intervals do not overlap, so there are no values of n that can satisfy this equation.

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Additional comment

The solutions would appear on the attached graph as points where the curves intersect above the x-axis. They do not intersect, hence the equation has zero solutions.

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If we were to solve this without regard to domain restrictions, we would square both sides to get ...

(n +2)² = -16 -5n

n² +9n +20 = 0 . . . . . put in standard form

(n +4)(n +5) = 0 . . . . . factor

n = {-5, -4} . . . . . . . . . both are extraneous solutions.

These "solutions" do not satisfy the requirement that the square root be positive.

Try it How many true, real number solutions does the equation n + 2 = √-16-5n have-example-1
User Loliki
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