Question

(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

Answer 1

`a) k=2`
`b) 1-i`
[Assuming `k in RR`]

Explanation:

`x^2 -kx +2=0`

In order to answer this question we must assume that `k` is real (i.e `k in RR`)

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 `(x-(1+i)) and (x-(1-i))` will be factors.

`:. (x-(1+i)) (x-(1-i)) =0`

`x^2-(1+i)x - (1-i)x +1^2+1^2 =0`

`x^2-x-ix-x+ix+2=0`

`x^2-x cancel(-ix)-xcancel(+ix)+2=0`

`x^2-2x+2 =0`

a) Equating coefficients `-> k=2`

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

Answer 2

(a) : `k=2,` (b) : `(1-i)` is the other root.

Explanation:

Part (a) :

Since `(1+i)` is a root of the eqn. `x^2-kx+2=0...(ast),` it must

satisfy the given eqn.

Hence, sub.ing `x=(1+i)` in `(ast),` we get,

`(1+i)^2-k(1+i)+2=0, i.e.,`

`1+2i+i^2-k(1+i)+2=0, or,`

`1+2i+(-1)-k(1+i)+2=0.`

`:. 2i+2=k(1+i).`

`:. (2(i+1))/(1+i)=k.`

`rArr k=2.`

Part (b) :

Sub.ing `k=2,` derived in Part (a), in `(ast),` we have,

`x^2-2x+2=0.`

`:. x^2-2x=-2.`

`"Completing the square, "x^2-2x+1=-2+1=-1,`

`:. (x-1)^2=i^2,`

`:. x-1=pmi,`

`:. x=1+pmi.`

Hence, the other root is, `(1-i).`