Multiple choice

The number of integral values of $a$ for which the quadratic equation $\left ( a+2 \right )x^{2}+2\left ( a+1 \right )x+a=0$ will have integer roots are

  1. $3$
  2. $2$
  3. $5$
  4. $4$
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D Correct answer
Explanation

For the equation (a+2)x^2 + 2(a+1)x + a = 0, the roots are x = [-2(a+1) +/- sqrt(4(a+1)^2 - 4a(a+2))] / 2(a+2). Simplifying the discriminant gives sqrt(4a^2 + 8a + 4 - 4a^2 - 8a) = sqrt(4) = 2. The roots are x = [-2a - 2 +/- 2] / 2(a+2), which are -1 and -a/(a+2). For -a/(a+2) to be an integer, a+2 must divide -a. Since -a = -(a+2) + 2, a+2 must divide 2. Possible values for a+2 are 1, -1, 2, -2, leading to a = -1, -3, 0, -4. Checking these, all yield integer roots.

AI explanation

For the equation (a+2)x^2 + 2(a+1)x + a = 0, the sum of the roots is -2(a+1)/(a+2) and the product is a/(a+2). If the roots are integers, both the sum and product must be integers, which means -2(a+1)/(a+2) = -2 + 2/(a+2) must be an integer. This restricts the possible values of a+2 to the integer divisors of 2: -2, -1, 1, and 2. Solving for a gives the values a = -4, a = -3, a = -1, and a = 0. Checking each shows they all yield valid quadratic equations with integer roots, meaning there are exactly 4 such integral values of a.

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