Question

Find a quadratic polynomial whose zeroes are 3and -4.​

Answer 1

Answer:

x2_x+4

hope will help u

Answer 2

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

Find the Quadratic polynomial whose zeroes are 3 and -4

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

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.

Now , we have

$\sf\:\alpha=3$

and $\sf\beta=-4$

Now ,

$\sf\blue{Sum of zeroes=\alpha+\beta}$

$\sf\implies\alpha+\beta=3+(-4)$

$\sf\implies\alpha+\beta=3-4$

$\sf\implies\alpha+\beta=-1....(1)$

and

$\sf\pink{Product of zeroes=\alpha\times\beta}$

$\sf\implies\alpha\times\beta=3\times(-4)$

$\sf\implies\alpha\times\beta=-12...(2)$

Thus,

The required Quadratic equation is

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

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

Then ,

$\sf\:f(x)=k[x^2-(-1)x+(-12)]$

$\sf\implies\:f(x)=k(x^2+x-12)$

Hence , the required Quadratic equation is

$\sf\purple{f(x)=k(x^2+x-12)}$

where k = any non- zero real number