Question

A quadratic polynomial has one of its zeros `1 + sqrt(5)` and it satisfies p(1) = 2. Find the quadratic polynomial

Answer 1

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)`