Question

How do I solve radical x^2-9 plus x+1=0

Answer 1

\[\sqrt{x ^{2}+9} +x +1=0\]

Answer 2

Now it's plus?

Answer 3

sorry, I wrote the question wrong

Answer 4

\(\sqrt{x^2} \sqrt{9} + x +1=0\)

Answer 5

would it be \[3\sqrt{x ^{2}}=-x-1\]

Answer 6

if the problem is
\[ \sqrt{x ^{2}+9} +x +1=0 \]
this is equivalent (add -(x+1) to both sides) to
\[ \sqrt{x ^{2}+9} = -(x +1) \]

square both sides
\[ x^2+9 = x^2 +2x +1 \\ 8=2x\\x=4\]

plug in x=4 into the original equation:

\[ \sqrt{16+9} +4 +1=0 \]
in these type problems, we take the positive square root
so this is
\[ 5 + 5=0 \\10=0\]

that means there is no solution

Answer 7

Thank you! @phi