Question

6
Find
quadric polynomial with zeroes (-2) and 1/2​

Answer 1

${\underline{\sf{To\:Find}}}$

Find the Quadratic polynomial whose zeroes are -2 and 1/2

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

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

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

${\underline{\sf{Answer}}}$

Let the zeroes of Quadratic equations be α and ß.

Thus ,

$\sf\:\alpha=-2$

and $\sf\beta=\dfrac{1}{2}$

Now sum of zeroes

$\sf=\green{\alpha+\beta}$

$\sf=-2+\dfrac{1}{2}$

$\sf=\dfrac{-4+1}{2}$

$\sf=\dfrac{-3}{2}....(1)$

and Product of zeroes

$\sf=\blue{\alpha\times\beta}$

$\sf=-2\times\dfrac{1}{2}$

$\sf=\dfrac{-2}{2}$

$\sf=-1...(2)$

Thus,

The required Quadratic equation is

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

Put the values of Equation (1)&(2)

Then ,

$\sf\:f(x)=k[x^2-\dfrac{(-3)}{2}x+(-1)]$

$\sf\implies\:f(x)=k(x^2+\dfrac{3}{2}x-1)$

$\sf\implies\:f(x)=k(\dfrac{2x^2+3x-1}{2})$

Hence , the required Quadratic equation is

$\sf\pink{f(x)=k(\dfrac{2x^2+3x-1}{2})}$

where k = any non- zero real number