Multiple choice

If the roots of the equation $(a+1)x^2-3ax+4a=0$ $(a\neq -1)$ are greater than unity, then the values of $a$ are.

  1. $\left[-\displaystyle\frac{16}{7}, -1\right)$
  2. $[0, -1]$
  3. $\displaystyle\left[-\frac{16}{7}, 1\right]$
  4. $[0, 1]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For roots of (a+1)x^2 - 3ax + 4a = 0 to be > 1: Let f(x) = (a+1)x^2 - 3ax + 4a. Conditions: D >= 0, (a+1)f(1) > 0, and -b/(2a) > 1. D = 9a^2 - 16a(a+1) = 9a^2 - 16a^2 - 16a = -7a^2 - 16a >= 0. This implies a(7a+16) <= 0, so a is in [-16/7, 0]. Checking f(1) = a+1 - 3a + 4a = 2a+1. (a+1)(2a+1) > 0. Vertex = 3a/(2(a+1)) > 1. Solving these yields the interval [-16/7, -1).

Keep practicing — related questions