### Instruction: Find a quadratic with real coefficients and quadratic term $x^2$ that has $5-4i$ as a root.### Response: I know that if a quadratic with real coefficients has a complex root, then its conjugate must also be a root.
So if $5-4i$ is a root, then $5+4i$ must also be a root.
To find the quadratic, I can use the fact that if $a$ and $b$ are roots of a quadratic, then the quadratic can be written as $(x-a)(x-b) = 0$.
So in this case, I have $(x-(5-4i))(x-(5+4i)) = 0$ as the quadratic equation.
To simplify this, I need to multiply out the brackets and collect the like terms.
I get $(x-5+4i)(x-5-4i) = x^2 - 10x + 25 - 16i^2 = x^2 - 10x + 41 = 0$.
This is the quadratic I was looking for.
It has real coefficients and quadratic term $x^2$, and $5-4i$ and $5+4i$ as roots.