Multiple choice

If the roots of the equation $x^{2}-2mx+m^{2}+1=0$ are imaginary, then $m$ lies in

  1. $(1, \infty)$
  2. $(-\infty, \infty)$
  3. $(-\infty, 1)$
  4. $All\ of\ these$
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B Correct answer
Explanation

For the roots of a quadratic equation to be imaginary, the discriminant (D = b^2 - 4ac) must be less than 0. Here, D = (-2m)^2 - 4(1)(m^2 + 1) = 4m^2 - 4m^2 - 4 = -4. Since -4 is always less than 0, the roots are imaginary for all real values of m.

AI explanation

The discriminant of the equation x^2 - 2mx + m^2 + 1 = 0 is D = (-2m)^2 - 4(1)(m^2 + 1). Expanding this gives D = 4m^2 - 4m^2 - 4, which simplifies to -4. Since the discriminant is strictly negative for all values of m, the roots are always imaginary, meaning m lies in the interval (-infinity, infinity).

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