Question

If alpha and beta are the zeros of the polynomial y 2 - 7y +2. find the quadratic polynomial whose zeros are a1/alpha and 1/beta

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{2y^{2}-7y+1=0}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given:}} \\ \tt: \implies {y}^{2}-7y+2= 0 \\ \\ \tt: \implies \alpha \: and \: \beta \: are \: zeroes \\ \\ \red{ \underline \bold{To \: Find:}} \\ \tt: \implies Polynomial\:whose\:zeroes\:are\:\frac{1}{\alpha}\:and\:\frac{1}{\beta} = ?$

• According to given question :

$\tt \circ \: {y}^{2} -7y +2 = 0 \\ \\ \tt \circ \: \alpha + \beta = \frac{ - b}{a} = {7} \\ \\ \tt \circ \: \alpha \beta = \frac{c}{a} = 2 \\ \\ \bold{For \: sum \: of \: zeroes}$

$\tt: \implies \frac{1}{ \alpha } + \frac{1}{ \beta } \\ \\ \tt: \implies \frac{ \beta + \alpha }{ \alpha \beta } \\ \\ \green{\tt: \implies \frac{7}{2} } \\ \\ \green{\tt \therefore \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{7}{2} } \\ \\ \bold{For \: product \: of \: zeroes} \\ \tt: \implies \frac{1}{ \alpha } \times \frac{1}{ \beta } \\ \\ \tt: \implies \frac{1}{ \alpha \beta } \\ \\ \tt: \green{\implies \frac{1}{2} } \\ \\ \green{\tt \therefore \frac{1}{ \alpha } \times \frac{1}{ \beta } = \frac{1}{2}} \\ \\ \bold{For \: Quadratic \: eqn} \\ \tt: \implies {y}^{2} - {(sum \: of \: zeroes})y + ( product \: of \: zeroes) = 0 \\ \\ \tt: \implies {y}^{2} - ( \frac{1}{ \alpha } + \frac{1}{ \beta } )y+ \frac{1}({ \alpha } \times \frac{1}{ \beta }) = 0 \\ \\ \tt: \implies {y}^{2} - \frac{7}{2}y + \frac{1}{2} = 0 \\ \\ \tt: \implies \frac{2 {y}^{2} - 7y + 1 }{2} = 0 \\ \\ \green{\tt: \implies 2 {y}^{2} - 7y + 1 = 0}$

Answer 2

Answer:

$\large\boxed{\sf{{y}^{2} - \dfrac{7}{2} y + \dfrac{1}{2} }}$

Step-by-step explanation:

Given a quadratic polynomial such that,

${y}^{2} - 7y + 2$

Also, it's zereos are alpha and beta.

Now, we know that,

Sum of roots = -b/a

Product of roots = c/a

Here, we have,

• a = 1
• b = -7
• c = 2

Therefore, we will get,

$= > \alpha + \beta = - ( \frac{ - 7}{1}) \\ \\ = > \alpha + \beta = 7$

And, also, we have,

$= > \alpha \beta = \frac{2}{1} \\ \\ = > \alpha \beta = 2$

Now, to find the quadratic polynomial having zeroes 1/alpha and 1/beta.

Sum of roots,

$= > \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ \alpha + \beta }{ \alpha \beta }$

Substituting the values, we get,

$= > \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{7}{2}$

Therefore, sum of roots = 7/2

Product of roots,

$= > \frac{1}{ \alpha } \times \frac{1}{ \beta } = \frac{1}{ \alpha \beta }$

Substituting the values, we get,

$= > \frac{1}{ \alpha } \times \frac{1}{ \beta } = \frac{1}{2}$

Now, we know that,

A quadratic polynomial can be written as,

$={y}^{2}-(sum\:of\:roots)y+(product\:of\:roots)$

Substituting the values, we get,

$= \bold{{y}^{2} - \dfrac{7}{2} y + \dfrac{1}{2} }$

Hence, the required polynomial is $\red{{y}^{2} - \dfrac{7}{2} y+ \dfrac{1}{2} }$