Multiple choice

If $a\in (-1, 1)$, then roots of the quadratic equation $(a-1)x^2+ax+\sqrt {1-a^2}=0$ are

  1. real

  2. imaginary

  3. both equal

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The discriminant D = b^2 - 4ac = a^2 - 4(a-1)(sqrt(1-a^2)). Since a is in (-1, 1), a^2 is positive and sqrt(1-a^2) is real. Analysis of the discriminant for values in the interval shows it is always non-negative, implying real roots.

AI explanation

To determine the nature of the roots, evaluate the discriminant of the quadratic equation, given by b squared minus 4ac. Substituting the coefficients gives the discriminant as a squared minus 4 times the quantity (a minus 1) times the square root of (1 minus a squared). Factoring out (1 minus a) from the term under a nested square root reveals the discriminant simplifies to (1 minus a) times (1 plus a minus 4 times the square root of (1 plus a)). Because a is between negative 1 and 1, both (1 minus a) and (1 plus a minus 4 times the square root of (1 plus a)) are always positive, meaning the discriminant is strictly positive. Therefore, the roots are real.

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