Question

if alpha and beta are the zeroes of x2+3x-10, find the polynomial of whose zeroes are 2 alpha and 2 beta​

Answer 1

Step-by-step explanation:

see that answer.......

Answer 2

Step-by-step explanation:

$\bf\underline{\underline{\red{Given:-}}}$

• $\rm \alpha \: and \: \beta \: are \: the \: zeroes \: of \: \: {x}^{2} + 3x - 10$

$\bf\underline{\underline{\blue{To\:Find:-}}}$

• $\rm The \: polynomial \: whose \: zeroes \: are \: 2 \alpha \: \: and \: \: 2 \beta$

$\bf\underline{\underline{\green{Solution:-}}}$

$\rm Given \: polynomial:- \: {x}^{2} + 3x - 10$

$\rm For \: a \: polynomial \: {ax}^{2} + bx + c$

$\rm Sum \: of \: zeroes = \dfrac{ - b}{ \: a}$

$\rm Product\: of \: zeroes = \dfrac{c}{ \: a}$

$\rm In \: {x}^{2} + 3x - 10$

$\leadsto \rm a = 1 \: , \: b = 3 \: \: and \: \: c = - 10$

$\leadsto \rm zeroes = \alpha \: \: and \: \: \beta$

$\rm Now:-$

$\rm Sum \: of \: zeroes \: (\alpha + \beta ) = \dfrac{ - b}{a} = \dfrac{ - 3}{1} = - 3$

$\rm Product\: of \: zeroes \: (\alpha \beta ) = \dfrac{ c}{a} = \dfrac{ - 10}{1} = - 10$

$\rm We\: Have :-$

• $\rm \alpha + \beta = -3$

• $\rm \alpha \beta = -10$

Now, we need to find the quadratic polynomial of zeroes

2α + 2β

$\rm Sum\: of \: zeroes:-$

$\rm = 2\alpha + 2\beta$

$\rm = 2(\alpha + \beta)$

$\rm = 2(-3)$

$\rm = -6$

$\rm Sum \: of \: zeroes = -6$

$\rm Product \: of \: zeroes :-$

$\rm = 2\alpha \times 2\beta$

$\rm = 2(\alpha\beta)$

$\rm = 2(-10)$

$\rm = -20$

$\rm Product \: of \: zeroes = -20$

$\rm The \: general \: form \: of \: quadratic \: polynomial \: is$

$\rm = {x}^{2} - ( sum \: of \: zeroes)x + product \: of \: zeroes$

$\rm = {x}^{2} - ( - 6)x + ( - 20)$

$\rm = {x}^{2} + 6x - 20$

$\underline{ \boxed{ \purple{ \rm \therefore \: Required \: polynomial = {x}^{2} + 6x - 20}}}$