Q1 Suppose a,b be complex nos. and x3+ax+b has a pair of complex conjugate roots then prove bot ha and b are real
Q2 If 0<a< b< c< d,then find location of roots of quad eqn f(x)=ax^2+[1-a(b+c)]x+abc-d=0 on number line
106Q1. sum of roots = -a = x1+x1+x2 = 2Re(x1) + x2
Let a = a1+ia2
b = b1+ib2
x1 = x+iy
x2 = c+id
then -a1 - ia2 = 2x + c+id
=> a2 = -d .. (i)
2x+c = -a1 .. (ii)
product of roots = -b = (x2+y2)(c+id)
=> -b1 = c(x2+y2)
and -b2 = d(x2+y2)
sum taking two at a time
0 = x2+y2 + (x1)x2 + (x2)(x1)
=> x2+y2 + 2Re(x1).x2 = 0
=> x2+y2 = -2x ..
and d=0
=> a2 = -d = 0
=> a = a1 and b=c So, a and b are all real
This is the usual approach .. im sure there must be something better
24hmm...vaise these both ques were in quad eqn chapter...so i think there must be an easy way ...
1x3+ax+b=0
roots are suppose p+iq,p-iq and r where p,q,r are real.
product is (p2+q2)r=-b/1 therefore bis real
pairwise product of roots is r(p+iq)+r(p-iq)+p2+q2=2pr+p2+q2=-a
thus a is also real
62sinchan your proof is almost correct.. not fully..
I dont know if you figured this out or wrote it by mistake.
but what is the guaranteee that the third root is a real number?
1sir the third root must be areal no.
because complex roots always occurs in conjugate pairs.
there cannot be 3 complex roots of any equation
62no sinchan that property is true only for a polynomial with real coefficients
but here you have to prove that!
1if the third root be x+iy then sum of roots is 0.
thus 2p+x+iy=0 where p,x,y are real hence y must be 0.
thus third root is real.
1Taking a,b as real coeffs , we can use Descartes sign change rule .
f(x) has no sign change , so no positive root .
f(-x) has one change , so at most one negative root.
Hence other two roots are complex conjugates .
But that is what is given.
So a and b have to be real.