Question

The sum and product of the zerosof the quardic polynomial are 1/2 and -1/2 respectively then the polynomial is

Answer 1

${\underline{\sf{Question}}}$

The sum and product of the zeros of the quardic polynomial are 1/2 and -1/2 respectively .Find the polynomial .

${\underline{\sf{Theory}}}$

if $\sf\alpha \: and \beta$ are the zeros of a quadratic polynomial f(x) . Then the polynomial f( x) is given by

$\sf \:f(x) = k(x {}^{2} - ( \alpha + \beta )x + \alpha \beta )$

or

$\sf \: f(x) = k(x {}^{2} - (sum \: of \: the \: zeroes)x + product \: of \: the \: zeroes)$

${\underline{\sf{Solution}}}$

Let $\sf \alpha \: and \: \beta$ are the zeros of the required polynomial.

It is given that $\sf \alpha + \beta = \dfrac{1}{2}$

$\sf \: and \: \alpha \beta = \dfrac{-1}{2}$

The quadratic polynomial is

$\sf \: f(x) =k( x {}^{2} - ( \alpha + \beta )x + \alpha \beta )$

$\sf \implies \: f(x) = k(x {}^{2} - \dfrac{1}{2}x + \dfrac{-1}{2})$

Taking the LCM:-

$\sf\implies\:f(x)=k(\dfrac{2x^{2}-x-1}{2})$

Answer 2

Answer :

2x² - x - 1

$\rule{100}2$

Solution :

It is given that,

• sum of zeroes ( α + β ) = 1/2
• product of zeroes ( αβ ) = -1/2

We have to find the polynomial.

We know that,

→ Quadratic Polynomial = k [ x² - ( sum of zeroes )x + ( product of zeroes )]

$\sf k [ x^2 - \dfrac{1}{2} \times x + \dfrac{(-1)}{2}]$

$\sf = k [x^2 - \dfrac{x}{2} - \dfrac{1}{2}]$

In order to remove 2 from the denominator, let k = 2.

Now,

$\sf = 2 [ x^2 - \dfrac{x}{2} - \dfrac{1}{2}]$

$\sf = 2x^2 - x - 1$

Hence,

the quadratic polynomial is 2x² - x - 1.

$\rule{100}2$