If $a, b, c$ are in $AP$ and one root of the equation $\displaystyle ax^{2}+bx+c=0$ is $2$, then the other root is
- $\displaystyle \frac{3}{4}$
- $\displaystyle \frac{-3}{4}$
- $\displaystyle -\frac{5}{4}$
- $\displaystyle \frac{-5}{2}$
Since a, b, and c are in AP, we have 2b = a + c. Substituting the root x = 2 into the quadratic equation ax^2 + bx + c = 0 gives 4a + 2b + c = 0, which simplifies to 5a + 2c = 0, or c/a = -5/2. Since the product of the roots is c/a, the other root r satisfies 2 * r = -5/2, which gives r = -5/4.
Since a, b, and c are in an arithmetic progression, we can use the property of an AP to write 2b = a + c. Substituting c = 2b - a into the given quadratic equation ax^2 + bx + c = 0 yields ax^2 + bx + 2b - a = 0. Because one root is 2, substituting x = 2 satisfies the equation, so 4a + 2b + 2b - a = 0, which simplifies to 3a + 4b = 0 and gives the ratio b/a = -3/4. Using the sum of roots formula where the sum equals -b/a, we have 2 + second root = -(-3/4) = 3/4. Solving for the second root gives 3/4 - 2 = -5/4. The result is -5/4.