Question

if alpha and bita are zeros of the quadratic polynomial p(x)=ax^2+bx+c, then evaluate alpha by bita + bita by alpha

Answer 1

Answer:

that is tge answer dear

Answer 2

$\bold{\huge{\underline{\underline{\rm{ Given :}}}}}$

$\alpha \: and \: \beta \: are \: the \: zeros \: of \\ \: polynomial \: a {x}^{2} + bx + c$

$\bold{\huge{\underline{\underline{\rm{ To\:Find :}}}}}$

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

$\rule{200}{1}$

Solution :-

We know that :-

• $sum \: of \: zeros \: = - \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$
• $product \: of \: zeros = \frac{contant \: term}{coefficient \: of \: {x}^{2} }$

In the given polynomial :-

Coefficient of x² = a

Coefficient of x = b

constant term = c

So,

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

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

Now come to The question :-

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

Using the identity :-

${x}^{2} + {y}^{2} = ( {x + y)}^{2} - 2xy$

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

Substituting the values of

$\alpha + \beta \: and \: \alpha \beta$

$= \frac{( - \frac{ {b} }{a})^{2} - 2 \times \frac{c}{a} }{ \frac{c}{a} } \\ \\ = \frac{ \frac{ {b}^{2} }{ {a}^{2} } - \frac{2c}{a} }{ \frac{c}{a} } \\ \\ = \frac{ \frac{ {b}^{2} - 2ac}{ {a}^{2} } }{ \frac{c}{a} } \\ \\ = \frac{ {b}^{2} - 2ac}{ {a}^{2} } \times \frac{a}{c} \\ \\ = \frac{ {b}^{2} - 2ac}{ac}$

$\rule{200}{1}$