Question

if
$\alpha and \: \beta$
are the zeroes of the polynomial f(x)=x
$x { }^{2} - 2x + 3$
then find the quadratic equation whose zeroes would be
$\alpha + 2 \: and \: \beta + 2$
​

Answer 1

Answer:

x² - 6x + 11

Step-by-step explanation:

Given---->

$\alpha \: and \: \beta \: are \: zeroes \: of \: the \: polynomial$

$f( \: x \: ) = {x}^{2} - 2x + 3$

To find---->

$quadratic \: equation \: whose \: zeroes \: are \: ( \alpha + 2) \: and \: ( \beta + 2 \: )$

Concept used---->

1)

$if \: polynomial \: ( \: a {x}^{2} \: + bx \: + c \: ) \: have \: zeroes \: \alpha \: and \: \beta \: then$

$\alpha \: + \: \beta = - \dfrac{b}{a}$

$\alpha \: \beta = \: \dfrac{c}{a}$

2)

$if \: we \: know \: the \: sum \: of \: zeroes \\ \: and \: product \: of \: zeroes \: then \: quadratic \: poynomial \: is \:$

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

Solution----> ATQ,

$f( \: x \: ) = {x}^{2} - 2x \: + 3$

$comparing \: it \: with \: a {x}^{2} + bx + c$

$a \: = 1$

$b \: = - 2$

$c \: = 3$

$\alpha \: + \: \beta = - \dfrac{ - 2}{1}$

$\alpha \: + \: \beta = 2$

$\alpha \beta = \dfrac{3}{1}$

$\alpha \beta = 3$

$now \: we \: have \: to \: find \: the \: polynomial \: whose \: zeroes \: are \: ( \alpha + 2) \: and \: ( \beta + 2)$

$now$

$sum \: of \: zeroes \: = ( \alpha + 2) + \: ( \beta + 2)$

$\: \: = \alpha + \beta + 4$

$\: \: = 2 + 4$

$\: \: = 6$

$product \: of \: zeroes \: = ( \alpha + 2) \: ( \beta + 2)$

$= \alpha \beta + 2 \alpha + 2 \beta + 4$

$= \alpha \beta + 2( \alpha + \beta ) + 4$

$= 3 + 2(2) + 4$

$= 3 + 4 + 4$

$= 11$

$now \: required \: polynomial \: is \:$

$x ^{2} - (sum \: of \: zeroes \: )x \: + product \: of \: zeroes \: = 0$

$={x}^{2} \: - (6)x + 11$

$= {x}^{2} \: - 6x \: + 11$