Let - - be the roots of x2-x-1-0 - If 1-a-1-b-1-c-2 - and 1-a-1-b-1-c-2 - then 1-a-1-1-b-1-1-c-1 is equal to

Roots of x^2 +x +1 = 0 are ω, ω^2 where ω is the cube root of unity. nbsp; Let α=ω, β=ω^2 Now, 1a +α+1b+α+1c +α = 2 β ⇒ nbsp;1a +ω+1b+ω+1c +ω = 2 ω nbsp; ...

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Byju's Answer

Standard XII

Mathematics

Properties of Cube Root of a Complex Number

Let α, β be t...

Question

Let $α, β$ be the roots of $x^{2} + x + 1 = 0$ . If $\frac{1}{a + α} + \frac{1}{b + α} + \frac{1}{c + α} = 2 β$ and $\frac{1}{a + β} + \frac{1}{b + β} + \frac{1}{c + β} = 2 α$ , then $\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}$ is equal to

\frac{2 α^{2}}{β}

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\frac{2 α}{β^{2}}

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\frac{α β}{2}

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2 α β

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Solution

The correct options are
A $\frac{2 α^{2}}{β}$
B $\frac{2 α}{β^{2}}$
D $2 α β$
Roots of $x^{2} + x + 1 = 0$ are $ω, ω^{2}$ where $ω$ is the cube root of unity.
Let $α = ω, β = ω^{2}$
Now,
$\frac{1}{a + α} + \frac{1}{b + α} + \frac{1}{c + α} = 2 β$
$\Rightarrow \frac{1}{a + ω} + \frac{1}{b + ω} + \frac{1}{c + ω} = \frac{2}{ω} [∵ ω^{3} = 1]$

$\frac{1}{a + β} + \frac{1}{b + β} + \frac{1}{c + β} = 2 α \Rightarrow \frac{1}{a + ω^{2}} + \frac{1}{b + ω^{2}} + \frac{1}{c + ω^{2}} = \frac{2}{ω^{2}}$

So, the equation
$\frac{1}{a + x} + \frac{1}{b + x} + \frac{1}{c + x} = \frac{2}{x} \dots (1)$
has roots $ω, ω^{2}$

Simplifying the equation, we get
$\Rightarrow x [(x + a) (x + c) + (x + a) (x + b) + (x + b) (x + c)] = 2 (x + a) (x + b) (x + c) \Rightarrow x [3 x^{2} + 2 (a + b + c) x + \dots] = 2 [x^{3} + (a + b + c) x^{2} + \dots] \Rightarrow x^{3} + 0 \cdot x^{2} + \dots = 0$
The roots of the cubic equations are $ω, ω^{2}, γ$ , so

Sum of roots
$ω + ω^{2} + γ = 0 \Rightarrow γ = 1 (∵ ω + ω^{2} = - 1)$

So, $x = 1$ is also a root of equation $(1)$ , putting it in equation $(1)$ , we get
$\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} = 2$
Now,
$α β = ω^{3} = 1 \frac{α^{2}}{β} = 1 \frac{α}{β^{2}} = 1$

Hence,
$\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} = 2 α β = \frac{2 α^{2}}{β} = \frac{2 α}{β^{2}}$

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Similar questions

Q. If

α, β

are roots of

x^{2} - p (x + 1) - c = 0

. Show that
i)

(α + 1) (β + 1) = 1 - c

&
ii)

\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + c} = 1

Q. If

a x^{2} + b x + c = 0

has roots

α

and

β

then-

a α^{2} + b α + c = 0

and

a β^{2} + b β + c = 0

α, β

are the roots of

x^{2} - p (x + 1) - c = 0, c \neq 1

, then

\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β 2 + 2 β + c} =

Q. If

α, β

are the roots of

a x^{2} + b x + c = 0

, then find the equation whose roots are

\frac{1}{a^{2}}, \frac{1}{β^{2}}

\frac{1}{a α + β}, \frac{1}{a β + b}

α + \frac{1}{β}, β + \frac{1}{α}

Q. If

(α + 1) (β - 1) + (β + 1) (α - 1) α + (α - 1) (β - 1) = 0

and

α (α + 1) (β + 1) - (α - 1) (β - 1) = 0

Also let

A = {\frac{α + 1}{α - 1}, \frac{β + 1}{β - 1}}

and

B = {\frac{2 α}{α - 1}, \frac{2 β}{β + 1}}

A \cap B \neq ϕ

, then find all the permissible value of parameter

" a "

Q. lf

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Let α,β be the roots of x2+x+1=0. If 1a+α+1b+α+1c+α=2β and 1a+β+1b+β+1c+β=2α, then 1a+1+1b+1+1c+1 is equal to

Let $α, β$ be the roots of $x^{2} + x + 1 = 0$ . If $\frac{1}{a + α} + \frac{1}{b + α} + \frac{1}{c + α} = 2 β$ and $\frac{1}{a + β} + \frac{1}{b + β} + \frac{1}{c + β} = 2 α$ , then $\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}$ is equal to