Question

If α and β are the zeroes of the quadratic polynomial 2x2 + 3x - 5, find the value of
1/α +1/β.


plx answer fassstttttt​

Answer 1

Step-by-step explanation:

Given -

α and β are Zeroes of polynomial 2x² + 3x - 5

To Find -

$\frac{1}{ \alpha} + \frac{1}{ \beta} \:$

Now,

According to the question

2x² + 3x - 5

2x² - 2x + 5x - 5

2x(x - 1) + 5(x - 1)

(2x + 5)(x - 1)

Zeroes of polynomial 2x² + 3x - 5 are

2x + 5 = 0 and x - 1 = 0

$\fbox \bold{x = \frac{ - 5}{2} \: \: \: and \: \: x = 1}$

$Let \: \bold{ \alpha} = 1 \\ and \\ \bold{ \beta} = \frac{ - 5}{2} \\ \\ Then, \\ The \: value \: of \: \frac{1}{ \alpha} + \frac{1}{ \beta} \: is \\ \\ \frac{1}{1} + \frac{1}{ \frac{ - 5}{2} } \\ = 1 - \frac{2}{5} \\ = \frac{5 - 2}{5} \\ = \frac{3}{5} \\ \\ Hence, \\ \fbox \bold{The \: value \: of \: \frac{1}{ \alpha} + \frac{1}{ \beta} \: is \: \frac{3}{5} }$

Answer 2

AnsWer :

3/5.

Concepts Required :

$\tt \blacksquare Sum \: of \: roots .\\ \tt \alpha + \beta = \frac{ - b}{a} = \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$\tt \blacksquare Product \: of \: roots .\\ \tt \alpha \times \beta = \frac{ c}{a} = \frac{constant \: term}{coefficient \: of \: {x}^{2} }$

To Find :

The value 1/α +1/β.

Solution :

We have equation,

$\tt2 {x}^{2} + 3x - 5. \\ \tt a = 2 \: , \: b = 3 \: , \: and, \: c = - 5.$

Now,

$\begin{aligned}\frac{1}{ \alpha } + \frac{1}{ \beta } & = \frac{ \alpha + \beta }{ \alpha . \beta } \\ & = \frac{ - 3}{\cancel2} \times \frac{\cancel2}{ - 5} \\ & = \frac{3}{5} \end{aligned}$

Therefore,the value of 1/α +1/β is 3/5.