Question

if alpha and beta be the zeros of the polynomial x²-1 then the value of 1/alpha + 1/beta is​

Answer 1

Answer:

0

Step-by-step explanation:

Given that alpha and beta are the zeros of the polynomial x² - 1. We need to find out the value of 1/alpha + 1/beta.

Polynomial: x² - 1 where a is 1, b is 0 and c is -1.

Sum of zeros = -b/a

alpha + beta = 0/1

alpha + beta = 0 --------------- (eq 1)

Product of zeros = c/a

alpha × beta = -1/1 ------------- (eq 2)

Now,

→ 1/alpha + 1/beta = (beta + alpha)/(alpha × beta)

Substitute the values in the above formula,

→ 1/alpha + 1/beta = 0/(-1/1)

→ 1/alpha + 1/beta = 0 × 1/(-1)

→ 1/alpha + 1/beta = 0

Therefore, the value of 1/alpha + 1/beta is 0.

Answer 2

$\displaystyle \sf \frac{1}{ \alpha } + \frac{1}{ \beta } \bf = 0$

Given :

α and β are the zeroes of the polynomial x² - 1

To find :

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

Concept :

If α and β are the zeroes of the quadratic polynomial ax² + bx + c , then

$\displaystyle \sf Sum \: of \: the \: zeroes = \alpha + \beta = - \frac{b}{a}$

$\displaystyle \sf Product \: of \: the \: zeroes = \alpha \beta = \frac{c}{a}$

Solution :

Step 1 of 3 :

Write down coefficients

Here the given polynomial is x² - 1

Comparing with the polynomial ax² + bx + c we get

a = 1 , b = 0 , c = - 1

Step 2 of 3 :

Find the value of α + β and αβ

$\displaystyle \sf Sum \: of \: the \: zeroes$

$\displaystyle \sf = \alpha + \beta$

$\displaystyle \sf = - \frac{b}{a}$

$\displaystyle \sf = - \frac{0}{1}$

$\displaystyle \sf =0$

$\displaystyle \sf Product \: of \: the \: zeroes$

$\displaystyle \sf = \alpha \beta$

$\displaystyle \sf = \frac{c}{a}$

$\displaystyle \sf = \frac{-1}{1}$

= - 1

Step 3 of 3 :

$\displaystyle \sf Find \: the \: value \: of \: \: \displaystyle \sf \frac{1}{ \alpha } + \frac{1}{ \beta }$

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

$\displaystyle \sf = \frac{ \alpha + \beta }{ \alpha \beta }$

$\displaystyle \sf = \frac{ 0 }{-1 }$

$= 0$

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