Question

alpha and beta are zeroes of the polynomial 2x^2 - 8x + 5 find the value of ( alpha + 1/beta ) x ( beta + 1/alpha ) [ answer fast please ]

Answer 1

Given Equation

$\sf\to 2x^2-8x+5$

To Find

$\to\sf\bigg(\alpha +\dfrac{1}{\beta } \bigg)\times\bigg(\beta +\dfrac{1}{\alpha } \bigg)$

Now Take

$\sf\to 2x^2-8x+5$

Compare with

$\sf ax^2+bx+c=0$

We get , a = 2 , b = -8 and c= 5

We Know that

$\sf\to\alpha +\beta =\dfrac{-b}{a}$

$\sf\to\alpha \beta =\dfrac{c}{a}$

So

$\sf\to\alpha +\beta =\dfrac{-(-8)}{2} = 4$

$\sf\to\alpha \beta =\dfrac{5}{2}$

Now simplify

$\to\sf\bigg(\alpha +\dfrac{1}{\beta } \bigg)\times\bigg(\beta +\dfrac{1}{\alpha } \bigg)$

$\sf\to\bigg(\alpha \beta +\dfrac{\alpha }{\alpha } +\dfrac{\beta }{\beta } +\dfrac{1}{\alpha \beta } \bigg)$

$\sf\to\bigg(\alpha \beta +1+1+\dfrac{1}{\alpha \beta } \bigg)$

$\sf\to\bigg(\alpha \beta +2+\dfrac{1}{\alpha \beta } \bigg)$

Now put the value

$\sf\to\bigg(\dfrac{5}{2} +2+\dfrac{1}{\dfrac{5}{2} } \bigg)$

$\sf\to\bigg(\dfrac{5}{2} +2+\dfrac{2}{{5}{} } \bigg)$

$\sf\to\dfrac{25+20+4}{10}$

$\sf\to \dfrac{49}{10}$

Answer

$\sf\to \dfrac{49}{10}$

Answer 2

Answer:

Step-by-step explanation: