Question

plz solve this sum..

Answer 1

Eqn
P(x) = X^3-6x^2+3x+10 = 0
a= 1, b= -6 , c= 3 , d= 10
Roots are given
Alpha = a
Beta = a+b
Gama = a+2b
Then according to zeroes of polynomial

Alpha+ beta + gama = -b/a= -(-6)
a+a+b+a+2b = 6
3a+3b = 6
a+b = 2 ____(1

alpha•beta•gama = -d/a
a(a+b)(a+2b) = -10
a(2)(a+2b) = -10 {since a+b =2}
2a(a+2b)= -10
a(a+2b) = -5 ____(2

alpha•beta+ beta•gama+gama•alpha = c/a = 3
a(a+b)+(a+b)(a+2b)+(a+2b)(a)= 3 ________(3)

From eqn(2) and (3)
2a + 2(a+2b) -5 =3
=> 2a-10/a-5 =3
2a^2 -10- 5a =3a
a^2-4a-5 = 0
a^2 -5a+a -5 = 0
a(a-5) + 1(a-5) = 0
(a-5)(a+1) = 0
Real value of a = 5

Hence put value of a in eqn (1)
a+b = 2
5+b = 2
b = 2-5 = -3

Answer 2

${x}^{3} - 6 {x}^{2} + 3x + 10 = 0 \\ {x}^{3} + {x}^{2} - 7 {x}^{2} - 7x + 10x + 10 = 0 \\ {x}^{2} (x + 1) - 7x(x + 1) + 10(x + 1) = 0 \\ (x + 1)( {x}^{2} - 7x + 10) = 0 \\ (x + 1)( {x}^{2} - 5x - 2x + 10) = 10 \\ (x + 1)(x - 5)(x - 2) = 0$
x1=-1, x2=2 and x3=5

as given in question x1=a ,x2=a+2b and x3=a+2b

so a=-1 and b = 3