# If$-i+3$ is a root of $x^{2}-6x+k=0$ then the value of $k$ is

$\begin{array}{1 1}(1)5&(2)\sqrt{5}\\(3)\sqrt{10}&(4)10\end{array}$

## 1 Answer

The given root is
$\alpha= -i+3$
The other root is
$\beta =i+3$
Sum $\alpha + \beta =-i +3 +i+3$
$\qquad= 6$
Product $\alpha \beta = (-i +3)(i+3)$
$\qquad= 3^2-i^2$
$\qquad= 9+1=10$
The quadratic equation is
$x^2$ -(sum of the roots)x + product of the roots=0
$x^2-6x+10=0$
comparing o the equation
$x^2-6x+k=0$
$k=10$
Hence 4 is the correct answer.
answered May 15, 2014 by

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