Question

If alpha and beta are the zeros of the quadratic polynomial f(x)= ax²+bx+c, then evaluate alpha/beta + beta/Alpha

Answer 1

Answer:

(b^2 - 2ac)÷(ac)

Step-by-step explanation:

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

$( \alpha \times \beta ) = \frac{ c}{a}$

From this question i.e.,

$\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha} = \frac{ \alpha {}^{2} + \beta {}^{2} }{ \alpha \times \beta }$

And we know,

$\alpha {}^{2} + { \beta }^{2} = ( \alpha + \beta )^{2}- (2 \times \alpha \times \beta)$

Putting the value of alpha^2 + beta^2 in the given eq. we get,

$\frac{b {}^{2} - 2ac }{a {}^{2} } \: \: \geqslant multiplying \: with \: \\ \frac{1}{ \alpha \times \beta } \: i.e. \frac{a}{c} \: \: we \: get \geqslant \\ = > \frac{b {}^{2} - 2ac }{ac}$

$\frac{b^{2} - 2ac}{ac} \: \: is \: the \: answer >$

Answer 2

Answer:

Step-by-step explanation:

( \alpha  +  \beta ) =   \frac{ - b}{a}  

( \alpha  \times  \beta ) =  \frac{  c}{a}  

From this question i.e.,

\frac{ \alpha }{ \beta }  +  \frac{  \beta  }{ \alpha}  = \frac{ \alpha  {}^{2} +  \beta  {}^{2}  }{ \alpha  \times  \beta }  

And we know,

\alpha  {}^{2}  +  { \beta }^{2}  = ( \alpha  +  \beta )^{2}- (2 \times  \alpha  \times  \beta) \\  

Putting the value of alpha^2 + beta^2 in the given eq. we get,

\frac{b {}^{2} - 2ac }{a {}^{2} } \:  \:  \geqslant  multiplying \: with \:  \\  \frac{1}{ \alpha  \times  \beta } \: i.e. \frac{a}{c}  \:  \: we \: get \geqslant \\  =  > \frac{b {}^{2} - 2ac }{ac}    

\frac{b^{2}  - 2ac}{ac} \:  \:  is \: the \: answer >