Question

If α,β are the zeros of a Quadratic polynomial such that α + β = 24, α- β = 8. Find a Quadratic polynomial having α and β as its zeros. *

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Quadratic\:equation\to x^{2}-24x+128=0}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: \implies \alpha + \beta = 24 \\ \\ \tt: \implies \alpha - \beta =8 \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Quadratic \: equation = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies \alpha + \beta = 24 - - - - - (1) \\ \\ \tt: \implies \alpha - \beta = 8 - - - - - (2) \\ \\ \text{Adding\: (1) \: from \:(2)} \\ \tt : \implies 2 \alpha = 24 + 8 \\ \\ \tt: \implies \alpha = \frac{32}{2} \\ \\ \green{\tt: \implies \alpha = 16} \\ \\ \tt: \implies 16 + \beta = 24 \\ \\ \tt: \implies \beta = 24 - 16 \\ \\ \green{\tt: \implies \beta = 8} \\ \\ \bold{For \: Quadratic \: equation :} \\ \tt: \implies {x}^{2} - ( \alpha + \beta )x + ( \alpha \beta ) = 0 \\ \\ \tt: \implies {x}^{2} - (24)x + (16 \times 8) = 0 \\ \\ \green{\tt: \implies {x}^{2} - 24x + 128 = 0}$

Answer 2

Answer:

Given :

$\alpha \: and \: \beta \: are \: the \: roots \: of \: quadratic \: polynomial.$

$\alpha + \beta = 24 \\ \alpha - \beta = 8$

To Find :

A quadratic polynomial having its zeros as$\alpha \: and \: \beta .$

Solution:

A quadratic polynomial has degree 2. And is expressed in the form ax²-(sum) x+(product).

According to the question,

$\alpha + \beta = 24 \\ \alpha = 24 - \beta \\ \\ = > \alpha - \beta = 8 \\ = > 24 - \beta - \beta = 8 \\ = > - 2 \beta = 8 - 24 \\ = > \beta = \frac{16}{2} = 8 \: \: and \: \: \alpha = (24 - 8) = 16$

Therefore,

$f(x) = {x}^{2} - ( \alpha + \beta )x + \alpha \beta \\ f(x) = {x}^{2} - 24x + 128$