Question

if alpha and beta are the zeroes of the polynomial
$x^{2} + 5x + 5$
,find the value of 1/alpha + 1/beta.

​

Answer 1

here alpha + beta =-b÷a

=-5÷1

alpha ×beta=5÷1

now 1÷alpha+1÷beta=alpha+beta ÷alpha×beta

=-5÷5=-1

=-1

Answer 2

$\underline\bold{\huge{SOLUTION\: :}}$

$x^{2} + 5x + 5$

Now let's find the sum of zeroes

$\alpha + \beta = -\frac{coefficient \: of \: x}{ciefficient \: of \: x^{2} }$

So, Here coefficient of $x^{2}$ is

1 and coefficient of x is 5.

$\alpha + \beta = \frac{-5}{1} =- 5$

Again their products will be

$\alpha \beta = \frac{constant \: term }{coefficient \: of \:x^{2} }$

So ,here Constant term is 5

and coefficient of $x^{2}$ is 1

$\alpha + \beta = \frac{5}{1} = 5$

Now Convert

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

In the form of sum and product.

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

Now put the value of sum of zeroes and

product of zeroes.

$\frac{1 }{ \alpha } + \frac{1}{ \beta } = \frac{sum \: of \: zeros}{product \: of \: zeroes}$

$\underline\bold{Replacing\: the\: value \:we\: got\: :}$

$\frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{-5}{5}$

$\frac{1}{ \alpha } + \frac{1}{ \beta } =- 1$

$\underline\bold{So, the\: value\: is\: -1\: :}$