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

Question

zyberg · Answer

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

princeharryyy · Answer

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

princeharryyy · Answer

answer would be 19 for them.

princeharryyy · Answer

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

zyberg · Answer

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

princeharryyy · Answer

:)