Question

If alpha and beta are the zeros of polynomial 3x2-2x-7 then find the value of alpha/beta + beta/Alpha

Answer 1

Given Polynomial:- 3x²-2x-7

Now..

Sum of zeroes=-b/a=-(-2)/3=2/3

Product of Zeroes=c/a=-7/3

ACCORDING TO THE QUESTION....

a/b+b/a

=)a²+b²/ab

= ( a + b)² - 2ab /ab

= ( 2/3)² - 2× -7/3 ÷ 7/3

= (4/9 + 14/3 ) ÷ 7/3

= 4 + 42/9 ÷ 7/3

= 46/9 ÷ 7/3

= 46/9 × 3/7

= 46/3 × 1/7

= 46/21

Answer 2

Answer:

$\frac{\alpha }{\beta } + \frac{\beta }{\alpha }$ =  $\frac{-46}{21}$

Step-by-step explanation:

If '$\alpha$' and '$\beta$'  are the roots of the equation  $ax^{2} +bx +c = 0$  

then we have,

Sum of roots = $\alpha +\beta$ = $\frac{-b}{a}$   and Product of roots = $\alpha \beta$ = $\frac{c}{a}$

Here, Given

If '$\alpha$' and '$\beta$'  are the roots of the equation  $3x^2-2x-7 = 0$

Comparing  $3x^2-2x-7 = 0$ with $ax^{2} +bx +c = 0$

we get,

a = 3 , b = -2 , c = -7

Then $\alpha +\beta$ = $\frac{-b}{a}$  =   $\frac{-(-2)}{3}$   = $\frac{2}{3}$ and $\alpha \beta$ = $\frac{c}{a}$ = $\frac{-7}{3}$

Required to find

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

Taking LCM as $\alpha \beta$ ,

 $\frac{\alpha }{\beta } + \frac{\beta }{\alpha }$ = $\frac{\alpha ^2+ \beta ^2}{\alpha \beta }$ -------------(1)

we know, the identity $a ^{2} + b ^2 = (a+b )^2 - 2ab$

Substitute the value of $\alpha ^2 + \beta ^2$ in equation (1) we get

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

= $\frac{( \frac{2}{3} )^2 - 2*(\frac{-7}{3}) }{\frac{-7}{3} }$

= $\frac{( \frac{4}{9} ) +(\frac{14}{3}) }{\frac{-7}{3} }$

= $\frac{4+42}{9} * (\frac{-3}{7} )$

= $\frac{46}{9} *\frac{-3}{7}$

= $\frac{-46}{21}$

Hence, $\frac{\alpha }{\beta } + \frac{\beta }{\alpha }$ =  $\frac{-46}{21}$