If a-d, a, a+d are the zeroes of p(x) = x³-3x²+x+1, find a and d.
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p(x) = x³-3x²+x+1
zeroes are a-d, a, a+d
sum of zeroes = -b/a
⇒ a-d + a + a+d = -(-3)/1
⇒ 3a = 3
⇒ a = 3/3 = 1
product of roots = -d/a
⇒ a(a-d)(a+d) = -1/1
⇒ 1(1-d)(1+d) = -1
⇒ 1 - d² = -1
⇒ d² = 1+1 = 2
⇒ d = √2 or -√2
So a = 1 and d = +√2 and -√2
zeroes are a-d, a, a+d
sum of zeroes = -b/a
⇒ a-d + a + a+d = -(-3)/1
⇒ 3a = 3
⇒ a = 3/3 = 1
product of roots = -d/a
⇒ a(a-d)(a+d) = -1/1
⇒ 1(1-d)(1+d) = -1
⇒ 1 - d² = -1
⇒ d² = 1+1 = 2
⇒ d = √2 or -√2
So a = 1 and d = +√2 and -√2
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