Question

Let $f\left(x\right)=x^2+kx;k$ is real number. The set of values of $k$ for which the equation $f\left(x\right)=0$ and $f\left(f\right(x\left)\right)=0$ have same real solution set.

Answer 1

$f\left(x\right)=x^2+kx=0$

$⟹x(x+k)=0$

$⟹x=0,-k$

These are real solution set of f(x).

$f\left(f\right(x\left)\right)=(x^2+kx{)}^2+k(x^2+kx)=0$

$⟹(x^2+kx)(x^2+kx+k)=0$

$⟹x^2+kx=0$ or $x^2+kx+k=0$

$x^2+kx=0$ gives two roots $0$ and $-k$ which are solution set of $f\left(x\right)$

Now, $x^2+kx+k=0$ should either satisfy these roots or should have no real roots since only these are real roots of $f\left(f\right(x\left)\right)$

for $x=0:$ $0+0+k=0⟹k=0$

for $x=-k:$ $k^2-k^2+k=0⟹k=0$

Hence for these two real roots, $k=0$.

If this equation has no real root, $D<0$

$⟹k^2-4k<0$

$⟹k(k-4)<0$

$⟹k\in (0,4)$

Hence final solution for k is $o\le k<4$ that is $k\in [0,4).$