polynomials question ! (2 Viewers)

[123j]

hey guys i got this question i couldnt figure out, could anyone help please?

Q) Two roots of the eqn.... x^3+ ax^2 + b = 0 .....are reciprocals of each other( a, b both are real)

a) show that the third root is equal to -b

b)show that a= b - 1/b

c)show that the two roots, which are reciprocals,will be real if - 1/2 < b<1/2



[ note: sign is less than or equal to]


help would be greatly appreciated, thanks. =]

 

[zeek]

The equation is a cubic polynomial .: it has 3 roots. Let the 3 roots be α,β,γ.
But one of the roots is equal to the reciporical of the other, .: the 3 roots must be α,1/α,γ

a) The product of the roots = -d/a
α.(1/α).γ=-b
.: γ=-b

b) The sum of the roots = -b/a
α +1/α + γ = -a
α +1/α -b= -a
.: α +1/α = b-a
Also, The sum of the roots two at a time = c/a
1+γ/α +αγ=0 {0 because the term x¹ doesn't exist}
1+γ(α +1/α) =0
1-b(b-a)=0
.: a=b-1/b {after rearranging}

c) The coeffecients, a and b, are real .: the polynomial must have real roots. From question b, we sub in the outer limits of -1/2<=b<=1/2 to judge the value of a.
a=-0.5--2 a=0.5-2
=1.5 =-1.5
.: -1/2<=a<=1/2
.: if b is real, as -1/2<=b<=1/2 then γ must be real as γ=-b, also, as a has the same restriction a is also real, .: α and 1/α must be real. Hence all roots are real.

 

[Yip]

The coeffecients, a and b, are real .: the polynomial must have real roots.

Im probably wrong, but isnt the argument that coefficients are real thus it has real roots invalid? consider x^2+x+1, it has real coefficients, but it obviously has complex roots...that argument only tells u there are conjugate roots isnt it....

i did (c) this way:

consider sum of roots of the cubic,
alpha+beta-b=-a, where alpha and beta are the reciprocal roots
alpha+beta=b-a=b-(b-1/b)=1/b
alphaxbeta=1 since they are reciprocals
From these 2 equations, it is obvious that they must satisfy the quadratic:
x^2-[1/b]x+1=0
To have real roots, which would mean alpha and beta would be real,
discriminant=[1/b^2]-4>=0

and thus we arrive at the condition, -1/2<=b<=1/2

 

[zeek]

i was assuming that complex roots wouldn't be considered because it's not in the 3 unit syllabus but yes your method is more correct

 

[123j]

oh i see thanks guys =]