(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

2 Answers
Alan N.
Oct 31, 2017

#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#

Ratnaker Mehta
Oct 31, 2017

(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).#

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