Multiple choice

If $\left( {2{x^2}\, - \,3x\, + \,1} \right)\left( {2{x^2}\, + \,5x\, + \,1} \right)\, = \,9{x^2}$, then equation has

  1. four real roots

  2. two real and two imaginary roots

  3. four imaginary roots

  4. None of the above
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A Correct answer
Explanation

The equation (2x^2 - 3x + 1)(2x^2 + 5x + 1) = 9x^2 can be rewritten by dividing by x^2: (2x - 3 + 1/x)(2x + 5 + 1/x) = 9. Let y = 2x + 1/x. Then (y - 3)(y + 5) = 9, so y^2 + 2y - 15 = 9, or y^2 + 2y - 24 = 0. Factoring gives (y + 6)(y - 4) = 0. For y = 4, 2x + 1/x = 4 implies 2x^2 - 4x + 1 = 0 (discriminant 16 - 8 = 8 > 0, two real roots). For y = -6, 2x + 1/x = -6 implies 2x^2 + 6x + 1 = 0 (discriminant 36 - 8 = 28 > 0, two real roots). Total is four real roots.

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