Question

If one of the roots of the equation $p{x}^2-6x+q=0$, where $p$ and $q$ are real numbers, is $\frac{3}{2}+\frac{7}{2}i$, then find the value of $|p-q|$.

Answer 1

Answer: 27

Since, $\frac{3}{2}+\frac{7}{2}i$ is one of the roots.
$\Rightarrow \frac{3}{2}-\frac{7}{2}i$ is the other root.
$p{x}^2-6x+q=0$
Sum of the roots $=\frac{-(-6)}{p}=\frac{6}{p}$
$\Rightarrow \left(\frac{3}{2}+\frac{7}{2}i\right)+\left(\frac{3}{2}-\frac{7}{2}i\right)=\frac{6}{p}$
$\Rightarrow p=2$
Product of the roots $=\frac{q}{p}$
$\Rightarrow \left(\frac{3}{2}+\frac{7}{2}i\right)\left(\frac{3}{2}-\frac{7}{2}i\right)=\frac{q}{p}$
$\Rightarrow \frac{9}{4}-\frac{49}{4}{i}^2=\frac{q}{2}$
$\Rightarrow \frac{9}{4}+\frac{49}{4}=\frac{q}{2}$
$\Rightarrow q=29$
$\Rightarrow |p-q|=|2-29|=27$