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 .....)
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.
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
Q1) To prove the relation 2b^3 -3abc +a^2d=0, given that the zeros of the polynomial f(x)=ax^3+3bx^2+3cx+d are in an arithmetic progression (A.P.), we need to express the roots in terms of an A.P. and then use Vieta's formulas to solve the equation.
Step 1: Express the roots in terms of an A.P.
Let the three zeros of the polynomial be (a - d), a, and (a + d), where 'a' is the common difference in the arithmetic progression, and 'd' is a constant.
Step 2: Use Vieta's formulas to relate the coefficients to the roots.
Vieta's formulas state that for a polynomial of degree n, the sum of the roots is equal to the negation of the coefficient of x^(n-1) divided by the coefficient of x^n, and the product of the roots is equal to the negation of the constant term divided by the coefficient of x^n.
For the given polynomial, f(x)=ax^3+3bx^2+3cx+d, we have:
Sum of the roots = (-3b) / a
Product of the roots = -d / a
Step 3: Use the expressions derived in Step 2 to solve for the given relation.
Using the expressions for the sum and product of the roots, we can rewrite the relation as:
2b^3 - 3abc + a^2d = 0
Substituting the expressions for the sum and product of the roots, we get:
2b^3 - 3a(-3b)(-d/a) + a^2d = 0
2b^3 + 9bd + a^2d = 0
Simplifying further, we obtain:
2b^3 + a^2d + 9bd = 0
Therefore, we have successfully proven the relation 2b^3 - 3abc + a^2d = 0.
Q2) To find the equation whose roots are (alpha + beta), (beta + gamma), and (gamma + alpha), given that alpha, beta, and gamma are the roots of the equation x^3 - 3x + 11 = 0, we can use Vieta's formulas.
Step 1: Express the given equation in terms of sums and products of roots.
The equation x^3 - 3x + 11 = 0 can be rewritten as:
x^3 + 0x^2 + (-3x) + 11 = 0
Here, the sum of the roots is 0 (according to Vieta's formulas), the product of the roots is (-11), and they satisfy the equation.
Step 2: Use Vieta's formulas to form the equation.
Let P(x) be the equation whose roots are (alpha + beta), (beta + gamma), and (gamma + alpha).
According to Vieta's formulas, the sum of the roots in P(x) is equal to the sum of the roots in the original equation, which is 0. Therefore, (alpha + beta) + (beta + gamma) + (gamma + alpha) = 0.
Also, the product of the roots in P(x) is equal to the product of the roots in the original equation, which is (-11). Therefore, (alpha + beta)(beta + gamma)(gamma + alpha) = -11.
Expanding the expression (alpha + beta)(beta + gamma)(gamma + alpha), we get:
(alpha*beta + alpha*gamma + beta*gamma + alpha^2 + beta^2 + gamma^2) = -11
This equation represents the polynomial P(x) whose roots are (alpha + beta), (beta + gamma), and (gamma + alpha).
Q3) To prove that c^2 + ac + b + 1 = 0, given that alpha and beta are the zeros of the cubic polynomial x^3 + ax^2 + bx + c, and alpha*beta = 1, we need to express the relation between the coefficients and the roots using Vieta's formulas and then substitute the given information to get the desired result.
Step 1: Use Vieta's formulas to relate the coefficients to the roots.
According to Vieta's formulas, for a polynomial of degree n, the sum of the roots is equal to the negation of the coefficient of x^(n-1) divided by the coefficient of x^n, and the product of the roots is equal to the negation of the constant term divided by the coefficient of x^n.
For the given polynomial x^3 + ax^2 + bx + c, we have:
Sum of the roots = (-a) / 1 = -a
Product of the roots = 1 / 1 = 1
Step 2: Use the expressions derived in Step 1 to solve for the given relation.
We know that the product of the roots, alpha*beta, is equal to 1. So, substituting alpha*beta = 1 into the expression derived from Vieta's formulas:
(alpha + beta) = -a
We can rewrite the relation c^2 + ac + b + 1 = 0 as:
(alpha + beta)^2 + a(alpha + beta) + b + 1 = 0
(-a)^2 + a(-a) + b + 1 = 0
a^2 - a^2 + b + 1 = 0
b + 1 = 0
Therefore, we have successfully proved that c^2 + ac + b + 1 = 0.