Question

if ∝,b are the zeros of f(x)=x square + x + 1, then find 1/∝+1/b. ​

Answer 1

ANSWER :–

• (1/∝) + (1/b) = -1

SOLUTION :–

GIVEN :–

• ∝, b are the zero's of polynomial f(x) = x² + x + 1.

TO FIND :–

• 1/∝+1/b = ?

SOLUTION :–

▪︎ We know that –

• Sum of roots = - (coffieciant of x) / (coffieciant of x²)

➨ ∝+ b = -(1)/(1)

➨ ∝+ b = - 1

• Product of roots = (constant term) / (coffieciant of x²)

➨ (∝)(b) = (1)/(1)

➨ (∝)(b) = 1

• Now Let's find –

= (1/∝) + (1/b)

= (∝+ b)/(∝.b)

• Put the values –

= (-1)/(1)

= -1

• Hence , (1/∝) + (1/b) = -1

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\dagger \: \: \: {\sf{ f(x)= {x}^{2} + x + 1 }}$

${\bf{\blue{\underline{To\Find:}}}}$

$\dagger \: \: \: {\sf{ \frac{1}{ \alpha} + \frac{1}{ \beta} =?}}$

${\bf{\blue{\underline{Now:}}}}$

$: \implies{\sf{ {x}^{2} + x + 1}}$

Compare it with ,

→ ax²+bx+c=0

$: \implies{\sf{ a = 1}}$

$: \implies{\sf{ b = 1}}$

$: \implies{\sf{ c = 1}}$

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$: \dagger \: \boxed{\sf{sum \: of \: roots \: = \alpha + \beta = \frac{ - b}{a} = \frac{ - (coeff. \: of \: x)}{coeff. \: of \: {x}^{2} } }}$

$: \implies{\sf{ \alpha + \beta = \frac{ - 1}{1} }}$

$: \implies{\sf{ \alpha + \beta = - 1}}$

__________________________________

$: \dagger \: \boxed{\sf{product \: of \: roots \: = \alpha \beta = \frac{ c}{a} = \frac{ constant \: term}{coeff. \: of \: {x}^{2} } }}$

$: \implies{\sf{ \alpha \beta = \frac{ 1}{1} }}$

$: \implies{\sf{ \alpha \beta = 1}}$

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$\dagger \: \: \underline {\mathfrak{ according \: to \: the \: ques:-}}$

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

$: \implies{\sf{ \frac{ \beta + \alpha}{ \alpha \beta} }}$

$: \implies{\sf{ \frac{ (1)}{ (1)} }}$

$: \implies{\sf{ -1}}$