# Is the conjugate of a solution to the time-independent Schroedinger equation also a solution?

Was reading the second chapter of Griffith's Introduction to Quantum Mechanics and have failed to understand why the conjugate of a solution to the Time-Indepedent Schrodinger Equation (hence TISE) is a solution itself.

It's at the end of Section 2.1 for reference (contained in Problem 2.1). I have the second edition.

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi.$$
$$-\frac{\hbar^2}{2m}\frac{d^2\psi^*}{dx^2} + V(x)\psi^* = E\psi^*,$$
since the potential and energy are clearly real, and the derivative is unaffected by conjugation. Therefore it is plain to see that if $\psi$ is a solution to the TISE, then so will its complex conjugate $\psi^*$.