It is given that $z=2+i$ is a root of $z^3+az^2-7z+15=0$, where $a$ is a real number - HSC - SSCE Mathematics Extension 2 - Question 6 - 2017 - Paper 1

Calculate $(2+i)^2$ and $(2+i)^3$:

$(2+i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i$

$(2+i)^3 = (2+i)(3 + 4i) = 6 + 8i + 3i + 4i^2 = 6 + 11i - 4 = 2 + 11i$

Substituting these back into the equation, we have: $(2 + 11i) + a(3 + 4i) - 7(2+i) + 15 = 0$

This simplifies to: $2 + 11i + a(3 + 4i) - 14 - 7i + 15 = 0$

Combining the real and imaginary parts:

• Real part: $2 - 14 + 15 + 3a = 3 + 3a$
• Imaginary part: $(11 - 7 + 4a)i = (4 + 4a)i$

Setting both parts to $0$ gives us two equations:

1. $3 + 3a = 0$
2. $4 + 4a = 0$

Solving the first equation: $3 + 3a = 0 \Rightarrow a = -1$

To confirm: From the imaginary part, $4 + 4a = 0 \Rightarrow a = -1$ also holds. Therefore, both equations are consistent.

Thus, the value of $a$ is: $a = -1$