# (a) Find k and (b) find a second solution for x^2 - kx + 2 = 0 where one solution is 1 + i?

## By the way, the "$i$" means imaginary number

Oct 31, 2017

a) k=2
b) 1-i
[Assuming $k \in \mathbb{R}$]

#### Explanation:

${x}^{2} - k x + 2 = 0$

In order to answer this question we must assume that $k$ is real (i.e $k \in \mathbb{R}$)

We are told the one root of the equation is $1 + i$. Since all of the coefficients are real the roots must occur as a conjugate pair.

Hence, we know that $\left(x - \left(1 + i\right)\right) \mathmr{and} \left(x - \left(1 - i\right)\right)$ will be factors.

$\therefore \left(x - \left(1 + i\right)\right) \left(x - \left(1 - i\right)\right) = 0$

${x}^{2} - \left(1 + i\right) x - \left(1 - i\right) x + {1}^{2} + {1}^{2} = 0$

${x}^{2} - x - i x - x + i x + 2 = 0$

${x}^{2} - x \cancel{- i x} - x \cancel{+ i x} + 2 = 0$

${x}^{2} - 2 x + 2 = 0$

a) Equating coefficients $\to k = 2$

b) The second root of the equation is: $1 - i$

Oct 31, 2017

(a) : $k = 2 ,$ (b) : $\left(1 - i\right)$ is the other root.

#### Explanation:

Part (a) :

Since $\left(1 + i\right)$ is a root of the eqn. ${x}^{2} - k x + 2 = 0. . . \left(\ast\right) ,$ it must

satisfy the given eqn.

Hence, sub.ing $x = \left(1 + i\right)$ in $\left(\ast\right) ,$ we get,

${\left(1 + i\right)}^{2} - k \left(1 + i\right) + 2 = 0 , i . e . ,$

$1 + 2 i + {i}^{2} - k \left(1 + i\right) + 2 = 0 , \mathmr{and} ,$

$1 + 2 i + \left(- 1\right) - k \left(1 + i\right) + 2 = 0.$

$\therefore 2 i + 2 = k \left(1 + i\right) .$

$\therefore \frac{2 \left(i + 1\right)}{1 + i} = k .$

$\Rightarrow k = 2.$

Part (b) :

Sub.ing $k = 2 ,$ derived in Part (a), in $\left(\ast\right) ,$ we have,

${x}^{2} - 2 x + 2 = 0.$

$\therefore {x}^{2} - 2 x = - 2.$

$\text{Completing the square, } {x}^{2} - 2 x + 1 = - 2 + 1 = - 1 ,$

$\therefore {\left(x - 1\right)}^{2} = {i}^{2} ,$

$\therefore x - 1 = \pm i ,$

$\therefore x = 1 + \pm i .$

Hence, the other root is, $\left(1 - i\right) .$