Question

please answer this question​

Answer 1

Given :

• α and β are the zeroes of the polynomial : x² - 5x + 6

To Find :

• The polynomial whose zeroes are : 1/α and 1/β

Solution :

Let's first find the roots of the equation : x² - 5x + 6

➣ $\mathtt{x^2\:-5x\:+\:6=0}$

➣ $\mathtt{x^2\:-\:3x\:-\:2x+6=0}$

➣ $\mathtt{x(x-3)\:\:-2(x-3)\:=0}$

➣ $\mathtt{(x-3)\:\:(x-2)\:=0}$

➣ $\mathtt{x-3=0\:\:or\:\:x-2=0}$

➣ $\mathtt{x=3\:\:or\:\:x=2}$

$\large{\boxed{\mathtt{\blue{Roots\:of\:the\:equation\:are\:2,3}}}}$

✪ Finding the polynomial :

Let,

• α = 2
• β = 3

Given that, the zeroes of the polynomial is 1/α and 1/β.

✪ In the zeroes of the polynomial, substitute appropriate value of α and β.

•°• Zeroes are :

• $\mathtt{\dfrac{1}{2}}$

• $\mathtt{\dfrac{1}{3}}$

✪ Sum of the zeroes :

α + β = $\mathtt{\dfrac{1}{2}\:+\:{\dfrac{1}{3}}}$

➣ $\mathtt{\dfrac{3+2}{6}}$

➣ $\mathtt{\dfrac{5}{6}}$ ____(1)

✪ Product of zeroes :

αβ = $\mathtt{\dfrac{1}{2}\:\times\:{\dfrac{1}{3}}}$

➣ $\mathtt{\dfrac{1}{6}}$ ____(2)

✪ We can form quadratic polynomial using,

• x² - (α + β)x + (αβ) = 0

From equation (1) and (2),

• a = 6
• b = 5
• c = 1

Therefore the required polynomial is :-

$\large{\boxed{\mathtt{\red{6x^2\:-\:5x\:+\:1\:=0}}}}$

Answer 2

AnswEr :

$\underline{\bigstar\:\textsf{According \: to \: given \: in \: question:}}$

$\:\bullet\:\sf\ Given \: polynomial : x^2 - 5x + 6$

$\normalsize\underline\mathfrak{Comaparing \: the \: polynomial :}$

$\normalsize\star\sf\ a = 1 \\ \normalsize\star\sf\ b = -5 \\ \normalsize\star\sf\ c = 6$

$\rule{100}1$

$\normalsize \underline\mathfrak{Sum \: of \: zeroes :}$

$\normalsize\ : \implies\sf\ \alpha + \beta = \frac{-b}{a}$

$\normalsize\ : \implies\sf\ \alpha + \beta = \frac{-(-5)}{1}$

$\normalsize\ : \implies\sf\alpha + \beta = 5$

$\normalsize\ : \implies{\boxed{\sf{\alpha + \beta = 5}}}$

$\normalsize \underline\mathfrak{Product \: of \: zeroes :}$

$\normalsize\ : \implies\sf\ \alpha \beta = \frac{c}{a}$

$\normalsize\ : \implies\sf\ \alpha \times\ \beta = \frac{6}{1}$

$\normalsize\ : \implies\sf\alpha \times\ \beta = 6$

$\normalsize\ : \implies{\boxed{\sf{\alpha\beta = 6}}}$

$\rule{100}1$

$\normalsize\underline\mathfrak{Now \: find \: the \: polynomial :}$

$\normalsize \underline\mathfrak{Sum \: of \: zeroes :}$

$\normalsize\ : \implies\sf\frac{1}{\alpha} + \frac{1}{\beta}$

$\normalsize\ : \implies\sf\frac{\alpha + \beta}{\alpha\beta} = \frac{5}{6}$

$\scriptsize\sf{\: \: \: \: \: \: \: \: \: [\red{Block \: the \: available \: values}]}$

$\normalsize\ : \implies{\boxed{\sf{\frac{1}{\alpha} + \frac{1}{\beta} = \frac{5}{6}}}}$

$\normalsize \underline\mathfrak{Product \: of \: zeroes :}$

$\normalsize\ : \implies\sf\frac{1}{\alpha} \times\ \frac{1}{\beta}$

$\scriptsize\sf{\: \: \: \: \: \: \: \: \: [\red{Block \: the \: available \: values}]}$

$\normalsize\ : \implies\sf\frac{1}{\alpha\beta} = \frac{1}{6}$

$\normalsize\ : \implies{\boxed{\sf{\frac{1}{\alpha\beta} = \frac{1}{6}}}}$

$\rule{100}1$

$\normalsize\star{\boxed{\sf{P(x) = k[x^2 - (\frac{1}{\alpha} + \frac{1}{\beta}) + \frac{1}{\alpha\beta}] }}}$

$\scriptsize\sf{\: \: \: \: \: \: \: \: \: [\purple{Block \: the \: available \: values}]}$

$\normalsize\ : \implies\sf\ k[x^2 - \frac{5}{6}x + \frac{1}{6}]$

$\normalsize\ : \implies\sf\ k[6x^2 - {5}x + 1]$

$\scriptsize\sf{\: \: \: \: \: \: \: \: \: [\purple{Here, \: k \: is \: constant, \: so \: k = 1 \: }]}$

$\normalsize\ : \implies\sf\ [6x^2 - {5}x + 1]$

$\normalsize\ : \implies{\underline{\boxed{\sf \green{P(x) = [6x^2 - {5}x + 1]}}}}$