$\frac{2}{1-3i}$ is a root of $px^2-4x+4=0$,where $p\in\mathbb{R}$. Find $p$ and range of $r$ such that $px^2-4x+4=rx^2$ has two distinct real roots.

Question

Questions

• Express $\frac{2}{1-3i}$ in the form of $a+ib$ where $a$ and $b$ are real numbers.

I solved it and get $a=\frac15$ and $b=\frac35$

• It is given that $\frac{2}{1-3i}$ is a root of the equation $px^{2}-4x+4=0$,where $p\in\mathbb{R}$.
  a. Find the value of $p$.
  b. Find the range of values of $r$ such that the quadratic equation $px^2-4x+4=rx^2$ has two distinct real roots.

Can anybody guide in how to proceed?

Answer 1

I assume that yoy want to find the value of $p$. First notice that $$z:=\frac{2}{1-3i}=\frac{2+6i}{10}=\frac{1+3i}{5}.$$ Now, since $z$ is a root of the polyonomial $p(x)=px^2-4x+4$, you need to solve for $p$ the following equation: $$p\left( \frac{1+3i}{5}\right)^2-4\left(\frac{1+3i}{5}\right)+4=0.$$ The solution is $p=10$.

Answer 2

Since $z_1 =\frac{1+3i}{5} $ and coefficients of polynomial are real we have $z_2 =\frac{1-3i}{5} $. So by Viete:

$$x_1+x_2 ={4\over p}\;\;\;\Longrightarrow \;\;\;\frac{2}{5}= {4\over p}\;\;\;\Longrightarrow \;\;\;p=10$$

Answer 3

A slightly different approach: $\frac12(1-3i)$ is a root of $z^2-z+\frac p4=0$. As $\frac12(1+3i)$ is the other one, then $\frac p4=\frac14(1-3i)(1+3i)$ etc.