A. Real solutions of \(x^n + k = 0\)
- Set all coefficients to zero except \( a \) and \( k \) (giving \( ax^5 + k = 0 \)). Count the x-intercepts as you vary \( a \) and \( k \).
- Set only \( b \) and \( k \) nonzero (degree 4). Which signs of \( k \) produce x-intercepts, and how many?
- Set only \( c \) and \( k \) nonzero (degree 3). Count the x-intercepts.
- Set only \( d \) and \( k \) nonzero (degree 2). Observe how the number of intercepts depends on \( k \).
- Set only \( e \) and \( k \) nonzero (degree 1). Observe the graph.
Conclusions (reveal after trying)
Odd \(n\) (degrees 1, 3, 5): the equation \( x^n + k = 0 \) has exactly 1 real solution for any real \( k \).
Even \(n\) (degrees 2, 4): the equation \( x^n + k = 0 \) has:
- 0 real solutions when \( k > 0 \),
- 1 real solution (\( x = 0 \)) when \( k = 0 \),
- 2 real solutions (\( x = \pm|k|^{1/n} \)) when \( k < 0 \).
Reason: even powers map both \(x\) and \(-x\) to the same positive value, so \(x^n \ge 0\) for even \(n\) — hence \(x^n = -k\) has real solutions only when \(-k \ge 0\).