Question

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

Answer 1

We are given the equation:

`n+2=\sqrt[]{-16-5n}`

Squaring both sides:

`(n+2)^2=-16-5n`

Expanding the square:

`n^2+4n+4=-16-5n`

Moving all terms to the left side:

`\begin{gathered} n^2+4n+4+16+5n=0 \\ \text{Simplifying:} \\ n^2+9n+20=0 \end{gathered}`

Factoring:

`(n+4)(n+5)=0`

This gives two possible solutions: n = -4 and n = -5.

Some equations, like the ones that have square roots, can produce extraneous solutions that satisfy the last equation, but not the original equation.

So we test both solutions in the original equation:

`\begin{gathered} -4+2=\sqrt[]{-16-5(-4)} \\ \text{Operating:} \\ -2=\sqrt[]{-16+20} \\ -2=2 \end{gathered}`

This is a false solution because the equation is false.

If we substitute the value x = -5, we get -3 = 3.

It's clear that none of the solutions is true, thus the answer is:

0 solutions

or

NO solutions