Question

Why 2x^2 + \sqrt{2x^2 + 1} = 11 roots are only +- 2? When I solve I get two more.

Answer 1

\(2x^2 + \sqrt{2x^2 + 1} = 11\)
\(2x^2 = n\)
\(n + \sqrt{n + 1} = 11\)
\((11 - n)^2 = (\sqrt{n+1})^2\)
\(n^2 - 22n + 121 = n + 1\)
\(n%2 - 23n + 120 = 0\)
\(D = 7^2\)
\(n_{1, 2} = \frac{23 +- 7}{2} = 8 and 15\)
So, n is 8 and 15, that would give the answer of x to be:
\(x = +- 2, +- \sqrt{7.5}\) Where am I wrong?

Answer 2

try plugging back the last two they won't satisfy the equation.

Answer 3

answer would be 19 for them.

Answer 4

and you are not wrong anywhere, one should check if the roots that came out in the final answers satisfy the equation above.

Answer 5

Oh! That's interesting! Thank you very much! :)

Answer 6

:)