If a and b are the roots of x^2+8x-5=0?

Question

find the value of a^3+b^3 and a^3b^3. Write the equetion whose roots are a^3 and b^3

Answer 1

Well, (x-a)(x-b)=x^2 +8x -5, so expanding gives ab=-5 and a+b=-8.
Now,
-512=(a+b)^3
=a^3+3a^2b+3ab^2+b^3
=a^3+b^3+3ab(a+b)
=a^3+b^3+120.
Thus,
a^3b^3=(ab)^3=-125 and
a^3+b^3=-632.
The polynomial with roots a^3 and b^3 will then be
(x-a^3)(x-b^3)=
x^2 -(a^3+b^3)x +a^3b^3
=x^2 +632x-125.

Answer 2

First, make sure that the coefficient of x^2 is 1.
divide the coefficient of x by two, and then square the result:
(8/2)^2 = 4^2 = 16
now add this number with +ve and -ve sign to the equation.

x^2 +8x -5= 0

x^2 +8x +(16 - 16) -5 = +16 -16

(x^2 +8x +16) -16 -5 =0
[ (x+4)^2 ] -21 =0
[ (x+4)^2 ] = 21

(x+4) = √21
x= 4 +or - √21
so a= 4+ √21, and b= 4 -√21

a^3 = (4+ √21)^3 = (4+ √21) (4+ √21) (4+ √21) =
(37 +8√21) (4 +√21) = 316 +69√21

b^3 = (4 -√21)^3 = (4 -√21) (4 -√21) (4 -√21) = 316 -69√21

a^3 + b^3 = 316 +316 +69√21 -69√21 = 632
a^3 b^3 = (316 +69√21) (316 -69√21) = (316^2) -(69√21)^2
= 99856 -99981 = -125

an equation with the two roots a^3 and b^3 is an equation where the sum of the two roots = the coefficient of x, and their products = the constant.
so the equation whose roots are a^3 and b^3 :

x^2 +632x -125=0

Answer 3

use the quadratic eq'n to get the roots a and b

then multiply (x-a^3)(x-b^3)

Answer 4

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Answer 5

shouldn't you be doing your own homework?

Answer 6

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