Question

9. If α and β are the zeros of the polynomial px2 + qx +r then find the values of a) 1/∝+ 1/β ii) α/β+ β/α

Answer 1

the answers of your question is in the pic which I attached with

Answer 2

Step-by-step explanation:

$\underline{given}$

$\alpha \: and \: \beta$

$\alpha \: and \: \beta are \: the \: zeroes \: of \: px {}^{2} - qx + r$

$\boxed{Tofind}$

$\boxed{(1/a+1/b)and(a/b+b/a}$

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

$a + b = \frac{ - ( - q)}{p}$

$\alpha + \beta = \frac{q}{p} .....(1)$

$\alpha \beta = \frac{c}{a}$

$\alpha \beta = \frac{r}{p} .....(2)$

now solving

$\underline{1/a+1/b}$

$\frac{ \alpha + \beta }{ \alpha \beta }$

putting the value of (1) and (2) we get

$= \frac{ \frac{q}{p} }{ \frac{r}{p} }$

$= \frac{q}{r}$

now solving

$\boxed{a/b+b/a}$

$\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha }$

$= \frac{ \alpha {}^{2} + \beta {}^{2} }{ \alpha \beta } ....(3)$

using (a+b)^2

$(a + b) {}^{2} = a {}^{2} + b {}^{2} + 2ab$

$(a + b) {}^{2} - 2ab = a {}^{2} + b {}^{2}$

• now using (1) and (2) we in the above identity

we get

$( \frac{q}{p} ) {}^{2} - 2( \frac{r}{p} ) = a {}^{2} + b {}^{2}$

$= \frac{q {}^{2} }{p {}^{2} } - \frac{2r}{p}$

$\frac{q {}^{2}p - 2p {}^{2} r }{p {}^{2} }$

$= \frac{p(q {}^{2} - 2pr) }{p {}^{2} }$

$= \frac{q { }^{2} - 2pr }{p} .....(4)$

now substituiting the value of (4) and (2) in (3) we get

$= \frac{ \frac{q {}^{2} - 2pr}{p} }{ \frac{r}{p} }$

$= \frac{q {}^{2} - 2pr }{r}$