NIMCET 2008 Mathematics PYQ — Let and be the roots of the equation . The equation whose roots a… | Mathem Solvex | Mathem Solvex
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NIMCET 2008 — Mathematics PYQ
NIMCET | Mathematics | 2008
Let α and β be the roots of the equation x2+x+1=0. The equation whose roots are α19 and β7 is:
Choose the correct answer:
A.
x2−x−1=0
B.
x2+x−1=0
C.
x2−x+1=0
D.
x2+x+1=0
(Correct Answer)
Correct Answer:
x2+x+1=0
Explanation
1. Identify the Roots
The given equation is x2+x+1=0. Using the quadratic formula, the roots are:
x=2(1)−1±12−4(1)(1)=2−1±−3=2−1±i3
These are the complex cube roots of unity, commonly denoted as ω and ω2.
So, let α=ω and β=ω2.
2. Properties of Cube Roots of Unity
Recall the fundamental properties of ω:
ω3=1
1+ω+ω2=0
3. Simplify the New Roots
We need to find the values of the new roots, α19 and β7:
For the first root:
α19=ω19
Since ω3=1, we can divide the power by 3 and look at the remainder: 19=(3×6)+1.
ω19=(ω3)6⋅ω1=(1)6⋅ω=ω
For the second root:
β7=(ω2)7=ω14
Dividing the power by 3: 14=(3×4)+2.
ω14=(ω3)4⋅ω2=(1)4⋅ω2=ω2
4. Form the New Equation
The new roots are simply ω and ω2 again.
Since the roots haven't changed in value (they just simplified back to the original roots), the equation remains the same.
The quadratic equation with roots ω and ω2 is:
(x−ω)(x−ω2)=0
x2−(ω+ω2)x+ω3=0
Using 1+ω+ω2=0⟹ω+ω2=−1 and ω3=1:
x2−(−1)x+1=0
x2+x+1=0
Conclusion
Because the high powers of the complex cube roots of unity cycle back to the original values, the resulting quadratic equation is identical to the starting one.
Correct Option: (d)
Explanation
1. Identify the Roots
The given equation is x2+x+1=0. Using the quadratic formula, the roots are:
x=2(1)−1±12−4(1)(1)=2−1±−3=2−1±i3
These are the complex cube roots of unity, commonly denoted as ω and ω2.
So, let α=ω and β=ω2.
2. Properties of Cube Roots of Unity
Recall the fundamental properties of ω:
ω3=1
1+ω+ω2=0
3. Simplify the New Roots
We need to find the values of the new roots, α19 and β7:
For the first root:
α19=ω19
Since ω3=1, we can divide the power by 3 and look at the remainder: 19=(3×6)+1.
ω19=(ω3)6⋅ω1=(1)6⋅ω=ω
For the second root:
β7=(ω2)7=ω14
Dividing the power by 3: 14=(3×4)+2.
ω14=(ω3)4⋅ω2=(1)4⋅ω2=ω2
4. Form the New Equation
The new roots are simply ω and ω2 again.
Since the roots haven't changed in value (they just simplified back to the original roots), the equation remains the same.
The quadratic equation with roots ω and ω2 is:
(x−ω)(x−ω2)=0
x2−(ω+ω2)x+ω3=0
Using 1+ω+ω2=0⟹ω+ω2=−1 and ω3=1:
x2−(−1)x+1=0
x2+x+1=0
Conclusion
Because the high powers of the complex cube roots of unity cycle back to the original values, the resulting quadratic equation is identical to the starting one.
