Multiple choice

If the quadratic equation $ax^2 + bx + b = 0$ has the integer roots, where a and b are also intergers, then which of the following cannot be true

  1. $4a-b=0$
  2. $4a+b=0$
  3. $b=0$
  4. $b\neq 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If roots are integers, the discriminant must be a perfect square. For ax^2 + bx + b = 0, D = b^2 - 4ab. If b = 0, x = 0 (integer). If b != 0, we test options. 4a+b=0 implies b = -4a, D = 16a^2 - 4a(-4a) = 32a^2, not a perfect square unless a=0.

AI explanation

Let the integer roots be r and s. By Vieta's formulas, r plus s equals -b/a and rs equals b/a. This means r+s = -rs, so r+s+rs = 0. Adding 1 to both sides gives rs + r + s + 1 = 1, which factors as (r+1)(s+1) = 1. The only integer solutions are r = s = 0, or one root is 0 and the other is -1. In either case b must be 0. Therefore, the condition 4a+b = 0 would require a = 0, which is impossible for a quadratic equation, making 4a+b = 0 impossible.

Keep practicing — related questions