If $a\in R^+$ and the roots of the equation $ax^2-3x+c=0$ are two consecutive odd positive integers then?
- $a\in \left(0, \dfrac{3}{4}\right]$
- $a\in (1, \infty)$
- $a\in (1, 4)$
- $a\in (0, \infty)$
Let the roots be n and n+2. The sum of roots is 3/a = 2n+2 and the product is c/a = n(n+2). For real roots, the discriminant D = 9 - 4ac >= 0, so ac <= 9/4. Given the roots are positive odd integers, the smallest possible roots are 1 and 3, so 3/a = 4 implies a = 3/4. As roots increase, a decreases, leading to the interval (0, 3/4].
Let the two consecutive odd positive integer roots be 2k minus 1 and 2k plus 1 for some positive integer k. The sum of the roots gives 4k divided by a equals 3, so a equals 4k divided by 3. The product of the roots gives c divided by a equals 4k squared minus 1, and substituting a equals 4k divided by 3 yields c equals (16k cubed minus 12k) divided by 9. The discriminant b squared minus 4ac must be strictly positive, which gives 9 minus 16k squared greater than 0, restricting k to the range of 0 to 0.75. This means a belongs to the interval from 0 to three-fourths.