Question

A quadratic polynomial has one of its zeros ${\displaystyle 1+\sqrt{5}}$ 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 ${\displaystyle 1+\sqrt{5}}$ is a zero of p(x)

∴ ${\displaystyle \text{a}{(1+\sqrt{5})}^2+\text{b}(1+\sqrt{5})+\text{c}}$ = 0

${\displaystyle \text{a}(1+5+2\sqrt{5})+\text{b}(1+\sqrt{5})+\text{c}}$ = 0

${\displaystyle 6\text{a}+2\text{a}\sqrt{5}+\text{b}+\text{b}\sqrt{5}+\text{c}}$ = 0  ...... (2)

If ${\displaystyle 1+\sqrt{5}}$ is zero then ${\displaystyle 1-\sqrt{5}}$ is also a zero of p(x).

∴ ${\displaystyle \text{a}{(1–\sqrt{5})}^2+\text{b}(1–\sqrt{5})+\text{c}}$ = 0

${\displaystyle \text{a}(1–2\sqrt{5}+5)+\text{b}(1–\sqrt{5})+\text{c}}$ = 0

${\displaystyle 6\text{a}–2\text{a}\sqrt{5}+\text{b}–\text{b}\sqrt{5}+\text{c}}$ = 0  ......(3)

(2) ⇒      ${\displaystyle 6\text{a} +2\text{a}\sqrt{5}+\text{b}+\text{b}\sqrt{5}+\text{c}}$ = 0
(1) ⇒          a  +      b   +     c                  = 2 
(2) – (1) ${\displaystyle 5\text{a} +2\text{a}\sqrt{5}+\text{b}\sqrt{5}}$             = – 2     ......(4)

(3) ⇒      ${\displaystyle 6\text{a} +2\text{a}\sqrt{5}+\text{b}-\text{b}\sqrt{5}+\text{c}}$ = 0
(1) ⇒          a  +      b   +     c                  = 2 
(3) – (1) ${\displaystyle 5\text{a} +2\text{a}\sqrt{5}-\text{b}\sqrt{5}}$             = – 2    ......(5)

(4) ⇒          ${\displaystyle 5\text{a} +2\text{a}\sqrt{5}+ \text{b}\sqrt{5}}$  = – 2
(5) ⇒          ${\displaystyle 5\text{a} +2\text{a}\sqrt{5}- \text{b}\sqrt{5}}$  = – 2
(4) + (5) ⇒ 10a  +     0     +     0     = – 4          ......(6)

a = ${\displaystyle -\frac{4}{10}=-\frac{2}{5}}$

Substituting the value of a in equation (4)

${\displaystyle 5\times -\frac{2}{5}+2\times -\frac{2}{5}\times \sqrt{5}+\text{b}\sqrt{5}}$ = – 2

${\displaystyle -2-\frac{4}{5}\sqrt{5}+\text{b}\sqrt{5}}$ = – 2

${\displaystyle \text{b}\sqrt{5}=\frac{4}{5}\sqrt{5}}$

b = ${\displaystyle \frac{4}{5}}$

Substituting the value of a and b in equation (1), we have

${\displaystyle -\frac{2}{5}+\frac{4}{5}+\text{c}}$ = 2

${\displaystyle \frac{-2+4}{5}+\text{c}}$ = 2

${\displaystyle \frac{2}{5}+\text{c}}$ = 2

c = ${\displaystyle 2-\frac{2}{5}}$

c = ${\displaystyle \frac{10-2}{5}}$

c = ${\displaystyle \frac{8}{5}}$

∴ The required quadratic polynomial is

p(x) = ${\displaystyle -\frac{2}{5}{x}^2+\frac{4}{5}x+\frac{8}{5}}$

p(x) =${\displaystyle -\frac{2}{5}(x^2-2x-4)}$