Ex 2.4, 3 (Optional) - Chapter 2 Class 10 Polynomials (Term 1)
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Ex 2.4, 3
If the zeroes of the polynomial x3 β 3x2 + x + 1 are a β b, a, a + b, find a and b.
For a cubic Polynomial
p(x) = ax3 + bx2 + cx + d
With zeroes Ξ±, π½ and Ξ³
We have
π + π½ + π = (βπ)/π
π"π½" + π½π + ππ = π/π
π"π½" π= (βπ
)/π
Now,
p(x) = x3 β 3x2 + x + 1
Comparing with p(x) = Ax3 + Bx2 + Cx + D,
A = 1, B = β3, C = 1 and D = 1
Zeroes are πΆ = a β b, π· = a and πΈ = a + b
Sum of zeroes
Sum of zeroes = (βπ΅)/π΄
πΆ + π· + πΈ = (βπ©)/π¨
a β b + a + a + b = 3
3a = 3
a = 1
Sum of Product zeroes
Sum of Product zeroes = πΆ/π΄
πΆπ·+ π· πΈ + πΈ πΆ = πͺ/π¨
(a β b)a + a(a + b) + (a + b) (a β b) = 1
a2 β ba + a2 + ab + a2 β b2 = 1
a2 + a2 + a2 βb2 = 1
3a2 β b2 = 1
Putting a = 1
3(1)2 β b2 = 1
3 β b2 = 1
3 β 1 = b2
b2 = 2
b = Β± βπ
Thus,
a = 1 and b = Β± βπ

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