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Question
A quadratic polynomial has one of its zeros `1 + sqrt(5)` and it satisfies p(1) = 2. Find the quadratic polynomial
Solution
Let p(x) = ax2 + bx + c be the required quadratic polynomial.
Given p (1) = 2 , we have
a × 12 + b × 1 + c = 2
a + b + c = 2 ......(1)
Also given `1 + sqrt(5)` is a zero of p(x)
∴ `"a"(1 + sqrt5)^2 + "b"(1 + sqrt5) + "c"` = 0
`"a"( 1 + 5 + 2sqrt5) + "b"(1 + sqrt5) + "c"` = 0
`6"a" + 2"a"sqrt(5) + "b" + "b"sqrt(5) + "c"` = 0 ...... (2)
If `1 + sqrt(5)` is zero then `1 - sqrt(5)` is also a zero of p(x).
∴ `"a"(1 – sqrt5)^2 + "b"(1 – sqrt5) + "c"` = 0
`"a"(1 – 2sqrt(5) + 5) + "b"(1 – sqrt5) + "c"` = 0
`6"a" – 2"a"sqrt(5) + "b" – "b"sqrt(5) + "c"` = 0 ......(3)
(2) ⇒ `6"a" + 2"a"sqrt(5) + "b" + "b"sqrt(5) + "c"` = 0
(1) ⇒ a + b + c = 2
(2) – (1) `5"a" + 2"a"sqrt(5) + "b"sqrt(5)` = – 2 ......(4)
(3) ⇒ `6"a" + 2"a"sqrt(5) + "b" - "b"sqrt(5) + "c"` = 0
(1) ⇒ a + b + c = 2
(3) – (1) `5"a" + 2"a"sqrt(5) - "b"sqrt(5)` = – 2 ......(5)
(4) ⇒ `5"a" + 2"a"sqrt(5) + "b"sqrt(5)` = – 2
(5) ⇒ `5"a" + 2"a"sqrt(5) - "b"sqrt(5)` = – 2
(4) + (5) ⇒ 10a + 0 + 0 = – 4 ......(6)
a = `- 4/10 = - 2/5`
Substituting the value of a in equation (4)
`5 xx - 2/5 + 2 xx - 2/5 xx sqrt(5) + "b"sqrt(5)` = – 2
`- 2- 4/5 sqrt(5) + "b"sqrt(5)` = – 2
`"b"sqrt(5) = 4/5 sqrt(5)`
b = `4/5`
Substituting the value of a and b in equation (1), we have
`-2/5 + 4/5 + "c"` = 2
`(- 2 + 4)/5 + "c"` = 2
`2/5 + "c"` = 2
c = `2 - 2/5`
c = `(10 - 2)/5`
c = `8/5`
∴ The required quadratic polynomial is
p(x) = `- 2/5x^2 + 4/5x + 8/5`
p(x) =`- 2/5 (x^2- 2x - 4)`
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