Q1). If the zeroes of he polynomial f(x)=ax^3+3bx^2+3cx+d are in A.P then show that 2b^3 -3abc +a^2d=0 (detailed solution .....)
Q2)if alpha, beta, gamma are the roots of equation x^3-3x+11=0 then find the equation whose roots are (alpha + beta),(beta + gamma),(gamma +alpha). (detailed solution ...)
Q3) let alpha , beta be the zeroes of the cubic polynomial x^3+ax^2+bx+c satisfying the relation alpha x beta=1.Prove that c^2+ac+b+1=0 (detailed solution .....)
3 answers
In Q1 would a long division (or synthetic division) give you the factors you need?
Don't know if you are familar with the properties of roots of a cubic, so for
ax^3 + bx^2 + cx + d = 0
the sum of the roots = -b/a
the product of the roots = -d/a
the sum of the products of roots taken in pairs = c/a
I will give the 2nd a try,
For yours: x^3 - 3x + 11 = 0 , a=1, b=0, c=-3 , d=11
let the roots be m, n, and p for easier typying instead of alpha, beta, and gamma
sum or roots = m+n+p = -0/1 = 0 or p = -m-n
product of roots = mnp = -11/1 = -11
sum of products taken two at a time
= mn + mp + np = 3/1 = 3
your new roots are m+n, m+p, and n+p
sum of new roots = 2m + 2n + 2p
= 2(m+n+p) = 0
product of new roots = (m+n)(m+p)(n+p) and since p = -m-n
= (m+n)(m-m-n)(n-m-n)
= (m+n)(-n)(-m)
= (-p)(-n)(-m) = -pmn = -(-11) = 11 , wow, that came out nice
sum of product in pairs
= (m+n)(m+p) + (m+n)(n+p) + (m+p)(n+p)
= m^2 + mp + mn + np + mn + mp + n^2 + np + mn + mp + np + p^2
= m^2 + n^2 + p^2 + 3mp + 3 mn + 3np
= m^2 + n^2 + p^2 + 3(m+n+p)
= m^2 + n^2 + p^2 + 3(0)
= m^2 + n^2 + p^2
now recall: (m+n+p)^2 = m^2 + n^2 + p^2 + 2mn + 2mp + 2np
= m^2+n^2+p^2 + 2(mn+mp+np)
= m^2+n^2+p^2 + 2(mn + m(-m-n) + n(-m-n))
= m^2+n^2+p^2 + 2(mn - m^2 - mn - mn - n^2)
= m^2+n^2+p^2 - 2(m^2 + n^2 + mn)
help me out here .....
I seem to be getting into some swamp here.
ax^3 + bx^2 + cx + d = 0
the sum of the roots = -b/a
the product of the roots = -d/a
the sum of the products of roots taken in pairs = c/a
I will give the 2nd a try,
For yours: x^3 - 3x + 11 = 0 , a=1, b=0, c=-3 , d=11
let the roots be m, n, and p for easier typying instead of alpha, beta, and gamma
sum or roots = m+n+p = -0/1 = 0 or p = -m-n
product of roots = mnp = -11/1 = -11
sum of products taken two at a time
= mn + mp + np = 3/1 = 3
your new roots are m+n, m+p, and n+p
sum of new roots = 2m + 2n + 2p
= 2(m+n+p) = 0
product of new roots = (m+n)(m+p)(n+p) and since p = -m-n
= (m+n)(m-m-n)(n-m-n)
= (m+n)(-n)(-m)
= (-p)(-n)(-m) = -pmn = -(-11) = 11 , wow, that came out nice
sum of product in pairs
= (m+n)(m+p) + (m+n)(n+p) + (m+p)(n+p)
= m^2 + mp + mn + np + mn + mp + n^2 + np + mn + mp + np + p^2
= m^2 + n^2 + p^2 + 3mp + 3 mn + 3np
= m^2 + n^2 + p^2 + 3(m+n+p)
= m^2 + n^2 + p^2 + 3(0)
= m^2 + n^2 + p^2
now recall: (m+n+p)^2 = m^2 + n^2 + p^2 + 2mn + 2mp + 2np
= m^2+n^2+p^2 + 2(mn+mp+np)
= m^2+n^2+p^2 + 2(mn + m(-m-n) + n(-m-n))
= m^2+n^2+p^2 + 2(mn - m^2 - mn - mn - n^2)
= m^2+n^2+p^2 - 2(m^2 + n^2 + mn)
help me out here .....
I seem to be getting into some swamp here.
hey, looking it over once again, I think I had it in the fourth last line
from::
now recall: (m+n+p)^2 = m^2 + n^2 + p^2 + 2mn + 2mp + 2np
m^2 + n^2 + p^2 = (m+n+p)^2 - 2mn - 2mp - 2np
= (m+n+p)^2 - 2(mn + mp + np)
= 0^2 - 2(3) = -6
so your new equation is
x^3 - 0x^2 + 6x + 11 = 0
x^3 + 6x + 11 = 0
check for any arithmetic errors
from::
now recall: (m+n+p)^2 = m^2 + n^2 + p^2 + 2mn + 2mp + 2np
m^2 + n^2 + p^2 = (m+n+p)^2 - 2mn - 2mp - 2np
= (m+n+p)^2 - 2(mn + mp + np)
= 0^2 - 2(3) = -6
so your new equation is
x^3 - 0x^2 + 6x + 11 = 0
x^3 + 6x + 11 = 0
check for any arithmetic errors
3 answers