Question

if alpha and beta are zeroes of the polynomial 3x^2+2x-6, find the value of (1) alpha - Berta
(2) alpha^2+ beeta^2
(3) alpha^3+beeta^3
(4) 1/alpha+ 1/ beeta

Answer 1

i am assuming alpha = a and beta = b for simplicity.

for a polynomial ax²+ bx +c,
sum of zeroes = -b/a
product of zeroes = c/a

Similarly for p(x)= 3x²+2x-6
a+b = -2/3
ab = -6/3 = -2

1. a-b:
(a-b)² = (a+b)²-4ab
= (-2/3)² -4(-2)
= 4/9 +8
= (4+9×8)/9
= 76/9
a-b = √76/3

2. a²+b² = (a+b)² -2ab
= (-2/3)² -2(-2)
= 4/9 +4
= 40/9
3. a³+b³ = (a+b)³ -3ab(a+b)
= (-2/3)³ -3(-2)(-2/3)
= -8/27 - 4
= -116/27
4. 1/a + 1/ b
= (a+b)/ab
= (-2/3)/(-2)
= 1/3

I hope you understand the approach.
Just play with the identities.
All the best!
And please check the calculations also.

Answer 2

Im not quite sure .. Cause im sleepy now but I hope that u atleast get the idea and knowledge for solving it....... ..
The last ans: i.e alpha cube + beeta cube...in that use the property of a^3+b^3........the formula is (a+b)(a^2-ab+b^2)....we know the value of a+ b( here it is alpha + beeta) and we also know the value of a^2+b^2..,.. Then we place the value of -ab and then simplify ...to get the answer......I could not write cause 5 files is the limit while i have 6..... The final answer will come : -116/27