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Common Roots of x³+2x²+2x+1 and x²⁰¹²+x²⁰¹⁴+1 | JEE

JEE Maths question with a full step-by-step solution.

Question

The number of common roots of the equations

x^3 + 2x^2 + 2x + 1 = 0

and

x^{2012} + x^{2014} + 1 = 0

1

2

correct

3

4

Solution

Step 1: Factorise the cubic. Test

x = -1

(-1)^3 + 2(-1)^2 + 2(-1) + 1 = -1 + 2 - 2 + 1 = 0

, so

(x + 1)

is a factor. Dividing:

x^3 + 2x^2 + 2x + 1 = (x + 1)(x^2 + x + 1)

The roots of the cubic are

x = -1

and the roots of

x^2 + x + 1 = 0

, namely the non-real cube roots of unity

\omega

and

\omega^2

(which satisfy

\omega^3 = 1

and

1 + \omega + \omega^2 = 0

). Step 2: Test

x = -1

in the second equation:

(-1)^{2012} + (-1)^{2014} + 1 = 1 + 1 + 1 = 3 \neq 0

x = -1

is not a common root. Step 3: Test

x = \omega

in the second equation. Reduce the exponents modulo

3

2012 = 3\cdot 670 + 2 \implies \omega^{2012} = \omega^2

2014 = 3\cdot 671 + 1 \implies \omega^{2014} = \omega^1 = \omega

Therefore:

\omega^{2012} + \omega^{2014} + 1 = \omega^2 + \omega + 1 = 0

Step 4: By the same computation (the exponent reductions are identical),

x = \omega^2

also satisfies the second equation. Step 5: Hence the common roots are

\omega

and

\omega^2

— exactly

2

common roots. Correct answer: (2)

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