Question

4. Alpha, beta, gemma are zeroes of cubic polynomial x^3-12x^2+44x+c. If alpha+gemma=2beta, find the value of c.

Answer 1

Hello,

see the attachment for answer ☺☺

________________________


Hope it helps u !!!

# Nikky

Answer 2

Note:

Here I am writing alpha as a, beta as b, and gamma as c.

Given f(x) = x^3 - 12x^2 + 44x + c.

Given that a + c = 2b----- (1)

We know that sum of the roots = -b/a

 a + b + c = -(-12)/1

                = 12  ------ (2)

ab + bc + ca = c/a

                     = 44/1

                     = 44    ------------- (3)

abc = -d/a

       = -c/1

       = -c   ------- (4)


From (1) & (4), we get

 = > 2b + b  = 12.

= > 3b = 12

= > b = 4   ------- (5)

Substitute b = 4 in (1), we get

= > a + c = 2b

= > a + c = 2(4)

= > a + c = 8.   ------- (6)

Substitute (5) in (4), we get

= > a(4) * c = -c

= > ac = -c/4    ------- (7)


On solving equation (3), we get

= > ab + bc + ca = 44

= > b(a + c) - c/4 = 44

= > b(8) - c/4 = 44

= > 32 - c/4 = 44

= > -c/4 = 44 - 32

= > -c/4 = 12

= > c = -48.



Therefore the value of c = -48



Hope this helps!