Question

If $\alpha \beta$ are the zeros of the polynomial $f(x)= x^{2}+x+1$, then $\frac{1}{\alpha} + \frac{1}{\beta}$=
(a) 1
(b) -1
(c) 0
(d) None of these

Answer 1

SOLUTION :

The correct option is (b) : - 1.

Given : α  and β are the zeroes of the  polynomial f(x) = x² + x +1

On comparing with ax² + bx + c,

a = 1, b= 1, c = 1

Sum of the zeroes = −coefficient of x / coefficient of x²

α + β  = -b/a = - 1/1 = - 1

α + β  = - 1 ………………….(1)

Product of the zeroes = constant term/ Coefficient of x²

αβ = c/a = 1/1 = 1

αβ = 1 ……………………(2)

The value of 1/α + 1/ β :

1/α + 1/ β = (α + β) /αβ

= - 1 / 1  

[From eq 1 & 2 ]

1/α + 1/ β = -1

Hence, the value of 1/α + 1/ β is -1.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Given :

f ( x ) = x² + x + 1

Comparing with a x² + b x + c

a = 1

b = 1

c = 1

We know that :

$\textsf{Sum of roots = }\mathsf{\frac{-b}{a}}$

$\textsf{Product of roots = }\mathsf{\frac{c}{a}}$

Thus :

$\alpha+\beta=\frac{-b}{a}\implies\frac{-1}{1}\implies-1$

$\alpha\times \beta=\frac{c}{a}\implies \frac{1}{1}\implies 1$

Hence the given value :

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

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

$\implies \mathsf{\frac{-1}{1}}$

$\implies -1$

ANSWER :

$\boxed{\textsf{OPTION B}}$

Hope it helps :-)

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