complex number question (1 Viewer)

[metalicarulez]

im havin trouble with these questions :

q1
3 - 2i is one root of x² + bx + c = 0 where b and c are real
Find b and c

q2
x² + 6x + k = 0 has one root @ where Im(@) = 2
If k is real find both roots of the equation and the value of k

q3
1 - 2i is one root of x² - (3 + i)x + k = 0. Find k and the other root of the equation

thanks a lot!

 

[spice girl]

all of these Q's use the rule that if the coefficients of a polynomial are real, then the complex solutions occur in conjugate pairs...

 

[sammeh]

so in q1 given that one root is 3 - 2i, the other must be 3 + 2i...?
which gives b = 6 and c = 13?

heh havent done complex polynomials yet, but no time like the present to learn!

continuing on, in q2 if one root has im(@) = 2, then the root @ = a + 2i, and the other root which i'll call & (degree 2 eqn, only 2 roots) = a - 2i.
solving for a by using @ + & = -b/1 = -6
@ + & = 2a = -6, a = -3
@ = -3 + 2i, & = -3 - 2i
and k = @& = (-3 + 2i)(-3 - 2i) = 13
so eqn is x² + 6x + 13 = 0 (which you would kno as soon as you find the roots, as they're the same as in q1...recycling questions, weak )

in q3 the roots wont occur in conjugate pairs, bcos the coefficiant of x¹ = (3 + i) which is not real, yah? which means its a simple "find the root" q, using same terminology as in q2;
@ = 1 - 2i
@ + & = 3 + i
& = 2 + 3i
k = @& = (1 - 2i)(2 + 3i) = 6 - i
eqn is x² - (3 + i)x + (6 - i) = 0

i sincerely doubt the validity of my answers, but if they're right, well i've learnt something today then

 

[:: ck ::]

all correct

last one is wrong i think

k = @& = (1 - 2i)(2 + 3i) = 6 - i
eqn is x2 - (3 + i)x + (6 - i) = 0

(1 - 2i)(2 + 3i) = (2 - (-6)) + i (3 - 4)
= 8 - i
not 6 - i

the rest u can do urself

 

[metalicarulez]

thanks guys

 

[Rahul]

Originally posted by metalicarulez
root of x² + bx + c

try the <> instead of []