Multiple choice

The value of $k$ for which the number $3$ lies between the roots of the equation $x^2+(1-2k)x+(k^2-k-2)=0$ is given by

  1. $2<k<5$
  2. $k<2$
  3. $2<k<3$
  4. $k>5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For 3 to lie between the roots of f(x) = x^2 + (1-2k)x + (k^2-k-2), f(3) must be < 0. f(3) = 9 + 3(1-2k) + k^2 - k - 2 = k^2 - 7k + 10 < 0. (k-2)(k-5) < 0, so 2 < k < 5.

AI explanation

Using the property that a number k lies between the roots of a quadratic if a*f(k) < 0, we substitute x = 3 into the given polynomial. The polynomial f(x) = x^2 + (1 - 2k)x + (k^2 - k - 2) evaluated at x = 3 gives 9 + 3(1 - 2k) + k^2 - k - 2. This simplifies to k^2 - 7k + 10. Since the coefficient of x^2 is positive, the condition is k^2 - 7k + 10 < 0. Factoring the inequality gives (k - 2)(k - 5) < 0, which means 2 < k < 5.

Keep practicing — related questions