Question

$\sf \ \alpha \ and \ \beta \ are \ the \ zeroes \ of \ the \ polynomial \ ax^2+bx+c \ , \ \\ \sf \: find \ a\ polynomial\ whose \ zeroes\ are \\ \sf \dfrac{1}{a\alpha+b}\ and\ \dfrac{1}{a\beta +b}$​

Answer 1

EXPLANATION.

$\sf \implies \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: polynomial \: = a {x}^{2} + bx + c \\ \\ \sf \implies \: sum \: of \: zeroes \: of \: the \: quadratic \: equation \: \\ \\ \sf \implies \: \alpha + \beta = \frac{ - b}{a} .........(1)\\ \\ \sf \implies \: products \: of \: zeroes \: of \: quadratic \: equation \\ \\ \sf \implies \: \alpha \beta = \frac{c}{a} ..........(2)$

$\sf \implies \: to \: find \: value \: of \: \dfrac{1}{a \alpha + b} \: + \: \dfrac{1}{a \beta + b}$

$\sf \implies \: \dfrac{1}{a \alpha + b} \: + \: \dfrac{1}{a \beta + b} \\ \\ \sf \implies \: \frac{a \beta + b + a \alpha + b}{(a \alpha + b)(a \beta + b)} \\ \\ \sf \implies \: \frac{a \alpha + a \beta + 2b}{( {a}^{2} \alpha \beta + a \alpha b + a \beta b + {b}^{2} ) } \\ \\ \sf \implies \: \frac{a( \alpha + \beta ) + 2b}{( {a}^{2} \alpha \beta + ab( \alpha + \beta ) + {b}^{2} ) }$

$\sf \implies \: \dfrac{a \times ( \dfrac{ - b}{a} ) + 2b}{( {a}^{2} \times \dfrac{c}{a} + ab \times ( \dfrac{ - b}{a}) + {b}^{2}) } \\ \\ \sf \implies \: \frac{ - b + 2b}{(ac + ( - {b}^{2}) + {b}^{2} ) } \\ \\ \sf \implies \: \frac{b}{ac}$

$\sf \implies \: product \: of \: zeroes \\ \\ \sf \implies \: \frac{1}{a \alpha + b} \: \times \: \frac{1}{a \beta + b} \\ \\ \sf \implies \: \frac{1}{(a \alpha + b)(a \beta + b)} \\ \\ \sf \implies \: \frac{1}{( {a}^{2} \alpha \beta + ab( \alpha + \beta ) + {b}^{2} ) }$

$\\ \\ \sf \implies \: \frac{1}{( {a}^{2} \times ( \dfrac{c}{a} ) + ab( \dfrac{ - b}{a} ) + {b}^{2}) } \\ \\ \sf \implies \: \frac{1}{ac - {b}^{2} + {b}^{2} } = \frac{1}{ac}$

Hence the required quadratic polynomial is $\\ \\ \sf \: \implies \: {x}^{2} - \frac{b}{ac}x \: + \frac{1}{ac} = 0$

Answer 2

Required Answer:-

Given:

• α and β are the zeros of the polynomial ax² + bx + c.

To find:

• The polynomial whose zeros are 1/(aα + b) and 1/(aβ + b)

Solution:

Here,

➡ α + β = -b/a

➡ αβ = c/a

Since, 1/(aα + b) and 1/(aβ + b) are the zeros of the polynomial,

Sum of zeros will be,

$\rm = \dfrac{1}{a \alpha + b} + \dfrac{1}{a \beta + b}$

$\rm = \dfrac{a \beta + b + a \alpha + b}{(a \alpha + b)(a \beta + b)}$

$\rm = \dfrac{a( \alpha + \beta ) + 2b}{(a \alpha + b)(a \beta + b)}$

$\rm = \dfrac{a( \alpha + \beta ) + 2b}{ {a}^{2} \alpha \beta + ab \beta + ab \alpha + {b}^{2}}$

$\rm = \dfrac{a( \alpha + \beta ) + 2b}{ {a}^{2} \alpha \beta + ab ( \alpha + \beta)+ {b}^{2}}$

Substitute the values. We get,

$\rm = \dfrac{a \bigg( \dfrac{ - b}{a} \bigg) + 2b}{ {a}^{2} \bigg( \dfrac{c}{a} \bigg) + ab \bigg( \dfrac{ - b}{a} \bigg)+ {b}^{2}}$

$\rm = \dfrac{ - b+ 2b}{ac - {b}^{2} + {b}^{2}}$

$\rm = \dfrac{b}{ac }$

Now, Product of zeros,

$\rm = \bigg( \dfrac{1}{a \alpha + b} \bigg) \bigg( \dfrac{1}{a \beta + b} \bigg)$

$\rm = \bigg( \dfrac{1}{ {a}^{2} \alpha \beta + ab( \alpha + \beta ) + {b}^{2} } \bigg)$

$\rm = \bigg( \dfrac{1}{ {a}^{2} \bigg( \dfrac{c}{a} \bigg) + ab \bigg( \dfrac{ - b}{a} \bigg) + {b}^{2} } \bigg)$

$\rm = \bigg( \dfrac{1}{ ac - {b}^{2} + {b}^{2} } \bigg)$

$\rm = \dfrac{1}{ac}$

Hence, the given quadratic polynomial will be,

$\rm {x}^{2} - \dfrac{b}{ac}x + \dfrac{1}{ac}$

which is our required answer.