Question

if alpha and beta are the zeroes of p(x) such that alpha +beta=24 and alpha-beta=8. Find quadratic polynomial having alpha and beta as its zeroes.​

Answer 1

Answer:

$\boxed{\huge{{x}^{2} - 24x + 128}}$

Step-by-step explanation:

$\bold{Given}\longrightarrow \\ \alpha \: and \: \beta \: are \: zeroes \: of \: p(x) \: such \: that \\ \alpha + \beta = 24 \\ \alpha - \beta = 8$

$\bold{To \: find}\longrightarrow \\ find \: p(x)$

$\bold{Concept \: used}\longrightarrow \\ 1)if \: we \: know \: the \: sum \: and \: product \: of \: zeroes \: then \\ polynomial \: is \\ {x}^{2} - (sum \: of \: zeroes)x + product \: of \: zeroes$

$\bold{Solution}\longrightarrow \\ \alpha + \beta = 24 ........(1)\\ \alpha - \beta = 8..........(2)$

$adding \: both \: equation$

$= > \alpha + \beta + \alpha - \beta = 24 + 8 \\ = > 2 \alpha = 32 \\ = > \alpha = \dfrac{32}{2} \\ = > \alpha = 16 \\ putting \: \alpha = 16 \: in \: (1) \\ = > 16 + \beta = 24 \\ = > \beta = 24 - 16 \\ = > \beta = 8 \\ now \\ \alpha + \beta = 24 \\ \alpha \beta = 16 \times 8 = 128 \\ required \: polynomial \: is \\ {x}^{2} - (sum \: of \: zeroes)x + product \: of \: zeroes \\ = > {x}^{2} - (24)x + 128 \\ = > {x}^{2} - 24x + 128$

Answer 2

Given :

$\rm{ \alpha \: and \: \beta \: are \: zeroes \: of \: p(x) \: in \: which} \\ \\ \alpha + \beta = 24 \\ \\ \alpha - \beta = 8$

To find:

$\rm{p(x)}$

Solution:

$\alpha + \beta = 24....(1) \\ \\ \alpha - \beta = 8....(2) \\ \\ \implies \: \alpha + \beta + \alpha - \beta = 24 + 8 \\ \\ \implies \: \: 2 \alpha = 32 \\ \\ \implies \alpha = \cancel \frac{32}{2} \\ \\ \implies \: \alpha = 16$

$\sf{putting \: ( \alpha ) = 16 \: in \: (p)} \\ \\ \implies \: 16 + \beta = 24 \\ \\ \implies \: \beta = 24 - 16 \\ \\ \implies \: \beta = 8 \\ \\ \\ \rm \: now : \\ \\ \alpha + \beta = 24 \\ \\ \implies \alpha \times \beta \: = 16 \times 8 = 128 \\ \\ \sf{required \: polynomial} \\ \\ {x}^{2} - ( {24)}^{x} + 128 \\ \\ = {x}^{2} - 24x + 128$