Question

If a, b, c are the roots of the equation x3-3x+11=0 then find the equation whose roots are (a+b), (b+c) and (c+a).

Answer 1

Given: A polynomial x^3 - 3x + 11 = 0, a, b, c are the roots.

To find: The equation whose roots are (a+b), (b+c) and (c+a).

Solution:

• Now we can write the equation as:

             x^3 + 0x^2 - 3x + 11.

• Let the roots be A, B, C

• Now, sum of zeroes is:

             A + B + C = 0/1 = 0 ............(i)

• Sum of the product of 2 zeroes:

            AB+BC+CA = -3/1 = -3

• Product of zeroes:

            A×B×C = -11/1 = -11

• Now when roota are A+B,B+C & C+A, then the value of root will be:

            A+B = -C or B+C = -A or A+C = -B .............from (i)

• Now, sum of zeroes of this polynomial :

           (-C)+(-A)+(-B) = -(A+B+C) = 0

• Sum of the product of 2 zeroes :

           (-A)(-B) + (-B)(-C) + (-C)(-A)  = AB + BC + CA = -3

• Product of its zeroes :

           (-A) x (-B) x (-C) = -ABC = -(-11) = 11

• So the polynomial will be:

           x^3 -(sum of its zeroes)x^2 + (sum of the product of its zeroes)x - (product of its zeroes)

           x^3 - 0x^2 + 3x - 11

           x^3 - 3x - 11

Answer:

      So the equation whose roots are (a+b), (b+c) and (c+a) is x^3 + 3x - 11.