Question

Fast friends tell me answer:If alpha and beta are the zeroes of 2x2-9x+10,form the polynomial whose zeroes are 1/alpha and 1/beta

Answer 1

Question :–

• If alpha and beta are the zeroes of 2x² - 9x + 10 = 0 , form the polynomial whose zeroes are 1/alpha and 1/beta.

ANSWER :–

GIVEN :–

• A quadratic equation 2x² - 9x + 10 = 0 which have two roots $\: \alpha \:$ and $\: \beta \:$

TO FIND :–

• Another quadratic equation whose zeroes are $\: \: \dfrac{1}{ \alpha } \:$ and $\: \dfrac{1}{\beta} \: \:$

SOLUTION :–

$\\ \: \implies \: { \bold{2 {x}^{2} - 9x + 10 = 0 \: }} \: \:$

• Splitting Middle term –

$\\ \: \implies \: { \bold{2 {x}^{2} - 5x - 4x+ 10 = 0 \: }} \: \:$

$\\ \: \implies \: { \bold{x(2 {x} - 5) - 2(2x - 5) = 0 \: }} \: \:$

$\\ \: \implies \: { \bold{(x - 2)(2 {x} - 5) = 0 \: }} \: \:$

$\\ \: \implies \: { \bold{x = 2 \: \: , \: \: \dfrac{5}{2} \: }} \: \:$

• Hence –

$\\ \: \implies \: { \bold{ \alpha = 2 \: \: , \: \: \beta = \dfrac{5}{2} \: }} \: \:$

• Now required equation –

$\\ \: \longrightarrow \: { \bold{ sum \: \: of \: \: roots = \dfrac{1}{ \alpha } + \dfrac{ 1}{ \beta } }} \: \:$

$\\ \: \implies \: { \bold{ sum \: \: of \: \: roots = \dfrac{1}{ 2} + \dfrac{ 1}{ ( \frac{5}{2} ) } }} \: \:$

$\\ \: \implies \: { \bold{ sum \: \: of \: \: roots = \dfrac{1}{ 2} + \dfrac{ 2}{ 5} }} \: \:$

$\\ \: \implies \: { \bold{ sum \: \: of \: \: roots = \dfrac{5 + 4}{ 2 \times 5} }} \: \:$

$\\ \: \implies \: { \bold{ sum \: \: of \: \: roots = \dfrac{9}{ 10} }} \: \:$

$\\ \: \longrightarrow \: { \bold{ Product \: \: of \: \: roots = \dfrac{1}{ \alpha } \times \dfrac{ 1}{ \beta } }} \: \:$

$\\ \: \implies\: { \bold{ Product \: \: of \: \: roots = \dfrac{1}{ \alpha \times \beta } }} \: \:$

$\\ \: \implies\: { \bold{ Product \: \: of \: \: roots = \dfrac{1}{ (2) ( \frac{5}{2} ) } }} \: \:$

$\\ \: \implies\: { \bold{ Product \: \: of \: \: roots = \dfrac{1}{ 5 } }} \: \:$

• We know that a quadratic equation –

$\\ \: \longrightarrow\: \large { \boxed{ \bold{ {x}^{2} - (sum \: \: of \: \: roots)x + (product \: \: of \: \: roots) = 0 }}} \: \:$

• So that , new equation –

$\\ \: \implies \: {{ \bold{ {x}^{2} - ( \frac{9}{10} )x + ( \frac{1}{5} ) = 0 }}} \: \:$

$\\ \: \longrightarrow\: \large { \boxed{ \bold{ 10 {x}^{2} - 9x + 2 = 0 }}} \: \:$

Answer 2

✯✯ QUESTION ✯✯

If alpha and beta are the zeroes of 2x²-9x+10,form the polynomial whose zeroes are 1/alpha and 1/beta..

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✰✰ ANSWER ✰✰

$\implies\tt{{2x}^{2}-9x+10=0}$

➥By Splitting the Middle Term : -

$\implies\tt{{2x}^{2}-(4x+5x)+10}$

$\implies\tt{{2x}^{2}-4x-5x+10}$

$\implies\tt{x(2x-5)-2(2x-5)}$

$\implies\tt{(x-2)(2x-5)}$

• x = 2
• x = 5/2

➥So , 2 and 5/2 are Zeroes of Polynominal 2x² - 9x + 10 ..

_______________________

★Sum Of Zeroes : -

$\implies\tt{\dfrac{1}{\alpha}+\dfrac{1}{\beta}}$

$\implies\tt{\dfrac{1}{2}+\dfrac{1}{(\frac{5}{2})}}$

$\implies\tt{\dfrac{1}{2}+\dfrac{2}{5}}$

$\implies\tt{\dfrac{5+4}{10}}$

${\implies\tt\dfrac{9}{10}}$

★Product Of Zeroes : -

$\implies\tt{\dfrac{1}{\alpha}\times{\dfrac{1}{\beta}}}$

$\implies\tt{\dfrac{1}{\alpha\times{\beta}}}$

$\implies\tt{\dfrac{1}{(2) (\frac{5}{2} )}}$

$\implies\tt{\dfrac{1}{5}}$

➥Now , Using Formula : -

$\implies\tt{{x}^{2}-(\alpha+\beta)x+(\alpha\beta)=0}$

➥Putting Values : -

$\implies\tt{{x}^{2}-\dfrac{9}{10}x+\dfrac{1}{5}=0}$

$\implies\tt{\large{\boxed{\bold{\bold{\red{\sf{{10x}^{2}-9x+2=0}}}}}}}$