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Question
If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b
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Solution 1
Since, (a - b), a, (a + b) are the zeroes of the polynomial x3 – 3x2 + x + 1.
Therefore, sum of the zeroes = (a - b) + a + (a + b) = -(-3)/1 = 3
⇒ 3a = 3 ⇒ a =1
∴ Sum of the products of is zeroes taken two at a time = a(a - b) + a(a + b) + (a + b) (a - b) =1/1 = 1
a2 - ab + a2 + ab + a2 - b2 = 1
⇒ 3a2 - b2 =1
Putting the value of a,
⇒ 3(1)2 - b2 = 1
⇒ 3 - b2 = 1
⇒ b2 = 2
⇒ b = ±`sqrt2`
Hence, a = 1 and b = ± `sqrt2`
Solution 2
p(x) = x3 - 3x2 + x + 1
Zeroes are a - b, a + a + b
Comparing the given polynomial with px3 + qx2 + rx + 1, we obtain
p = 1, q = - 3, r = 1, t = 1
Sum of zeroes = a - b + a + a +b
`(-q)/p = 3a`
`(-(-3))/1 =3a`
3 = 3a
a = 1
The zeroes are 1-b, 1, 1+b
Multiplication of zeroes = 1(1-b)(1+b)
`(-t)/p=1-b^2`
`(-1)/1 = 1 - b^2`
1- b2 = -1
1 + 1 = b2
b = ±`sqrt2`
Hence a = 1 and b = `sqrt2` or - `sqrt2`
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