Question

$\large\boxed\colour{purple}{Question:-}$

Without actually calculating the zeroes, form a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial $5x²+2x–3$

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Answer 1

Answer:

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.

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Answer 2

Answer:

${ \sf{Given \: Quadratic \: Polynomial : 5 {x}^{2} + 2x - 3}}$

• We have to find the quadratic polynomial whose zeroes Are reciprocal of given quadratic polynomial

• Let us find sum and product of zeroes for given polynomial

We know that,

• ${{ \sf{ \alpha + \beta = \frac{ - b}{a } }}}$
• ${ { \sf{ \alpha \beta = \frac{c}{a}}}}$

$\: : { \implies{ \sf{ \alpha + \beta = \frac{ - b}{a } = \frac{ - 2}{5} }}} \\ \\ : { \implies{ \sf{ \alpha \beta = \frac{c}{a} = \frac{ - 3}{5} }}}$

• We had finded the zeroes. Now we have to reciprocal of the following zeroes.

$: { \implies{ \sf{ \frac{1}{ \alpha} + \frac{1}{ \beta} = \frac{ \alpha + \beta}{ \alpha \beta} = \frac{2}{3} }}} \\ \\ : { \implies{ \sf{ \frac{1}{ \alpha \beta } = \frac{1}{ \frac{ - 3}{5} } = \frac{ - 5}{3}}}}$

Finding the quadratic polynomial:-

${ \boxed{ \sf{</p><p>General \: Form : {x}^{2} - ( \alpha + \beta)x + \alpha \beta }}}$

$: { \implies{ \sf{k \bigg[ {x}^{2} - \bigg( \frac{2}{3} \bigg)x + \bigg( \frac{ - 5}{3} \bigg) \bigg]}}} \\ \\ : { \implies{ \sf{k \bigg[ {x}^{2} - \frac{2}{3} x - \frac{5}{3} \bigg ]}}} \\ \\ : { \implies{ \sf{k \bigg[ \frac{3 {x}^{2} - 2x - 5 }{3} \bigg ]}}} \\ \\ taking \: k = 3 \\ \\ : { \implies{ \sf{3 \bigg[ \frac{3 {x}^{2} - 2x - 5}{3} \bigg]}}} \\ \\ : { \implies{ \sf{3 {x}^{2} - 2x - 5}}}$

Therefore,

• 3x² - 2x - 5 is the required polynomial