Question

$\huge\mathbb\fcolorbox{Purple}{pink}{⭐question⭐}$
Without actually calculating the zeroes, form a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial 5x²+2x–35x²+2x–3 ​

Answer 1

●$\large\sf \bf \underline\purple{ your\: \: answer \: !!}$

Without actually calculating the zeroes,form a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial 5x2+2x-3 is 3x^2 - 2x - 5.

Answer 2

Answer:

$★\text{Let } \alpha \text{ and \: } \beta \text{ \: are roots of}$

$⇒ {5x}^{2} + 2x - 3$

$⇒ \alpha + \beta = - \frac{2}{5} \sf \colorbox{god} {and} \: \alpha \beta = - \frac{3}{5}$

★ Now let new polynomial has roots = A & B

$⇒ATQ \: A = \frac{1}{ \alpha } \:\sf \colorbox{god} {and \: } B = \frac{1}{ \beta }$

$★\sf \colorbox{god} {Now sum of roots}$

$⇒A + B = \frac{1}{ \alpha } + \frac{1}{ \beta }$

$⇒ \frac{ \beta + \alpha }{ \alpha \beta } = - \frac{ \frac{2}{5} }{ - \frac{3}{5} } = \frac{2}{3}$

$★\sf \colorbox{god} {And Product of roots}$

$⇒A \times B = \frac{1}{ \alpha } \times \frac{1}{ \beta }$

$⇒ \frac{1}{ \alpha \beta } = \frac{1}{ - \frac{3}{5} } = - \frac{5}{3}$

$\textbf{Polynomial Will be}$

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

$⇒ {x}^{2} - \frac{2x}{3} + \left( - \frac{5}{3} \right) \] = 0$

$⇒ {3x}^{2} - 2x - 5 = 0$

Hence,

$= {3x}^{2} - 2x - 5 = 0$

$\\ \\ \\ \\ \sf \colorbox{lightgreen} {\red★ANSWER ᵇʸɴᴀᴡᴀʙ}$