Question

If a, b,y are the zeroes of the cubic polynomial 2x3 - 3x2 + 6x + 1 then find the value of
4α²+ 4β² + 4gama²​

Answer 1

Answer:

-15

Step-by-step explanation:

(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)...(i)

Given equation:

2x3 - 3x2 + 6x + 1 (a,b,c--are the roots)

Comparing the equation with

Ax3 +Bx2 +Cx+D,we get,

A=2

B=-3

C=6

D=1

using the formula of roots--(here)

(a+b+c)=-B/A= -(-3)/2=3/2

(ab+bc+ca)=C/A=6/2=3

putting this values in equation (i):

(3/2)^2=a^2+b^2+c^2+2(3)

9/4=a^2+b^2+c^2+6

a^2+b^2+c^2=(9/4)-6= -15/4....(ii)

WE NEED:

4a^2+4b^2+4c^2

=4(a^2+b^2+c^2)

Putting value from equation (ii) we get:

=4(-15/4)= -15(ANS)