If - are the distinct roots of the equation x2-x-1-0- then -101-107 is equal to

Given: x^2 x+1=0 ⇒ x=1±√(1 4)2=1± i√(3)2 ⇒ ω^2=1+i√(3)2=α (say) and ω=1 i√(3)2=β (say) Now, nbsp; α^101+β^107 =( ω^2)^101+( ω)^107 = ω^202 ω^107 = ((ω^3)^67 ...

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Byju's Answer

Standard XII

Mathematics

Range of Quadratic Expression

If α,β∈ℂ are ...

Question

If $α, β \in C$ are the distinct roots of the equation $x^{2} - x + 1 = 0,$ then $α^{101} + β^{107}$ is equal to

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Solution

The correct option is D $1$
Given: $x^{2} - x + 1 = 0$
$\Rightarrow x = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm i \sqrt{3}}{2}$
$\Rightarrow - ω^{2} = \frac{1 + i \sqrt{3}}{2} = α$ (say)
and $- ω = \frac{1 - i \sqrt{3}}{2} = β$ (say)
Now,
$α^{101} + β^{107}$
$= (- ω^{2})^{101} + (- ω)^{107}$
$= - ω^{202} - ω^{107}$
$= - ((ω^{3})^{67} \cdot ω + (ω^{3})^{35} \cdot ω^{2})$
$= - (ω + ω^{2}) = 1$

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If α,β∈C are the distinct roots of the equation x2−x+1=0, then α101+β107 is equal to

The correct option is D 1Given: x2−x+1=0 ⇒x=1±√1−42=1±i√32 ⇒−ω2=1+i√32=α (say) and −ω=1−i√32=β (say) Now, α101+β107 =(−ω2)101+(−ω)107 =−ω202−ω107 =−((ω3)67⋅ω+(ω3)35⋅ω2) =−(ω+ω2)=1

If $α, β \in C$ are the distinct roots of the equation $x^{2} - x + 1 = 0,$ then $α^{101} + β^{107}$ is equal to