If a and b are the zeros of the quadratic polynomial $f(x) = 6{x^2} + x - 2$, find the value of $\frac{a}{b} + \frac{b}{a}$.
- $\dfrac12$
- $\dfrac32$
- $\dfrac52$
- $-\dfrac{25} {12} $
For 6x^2 + x - 2 = 0, sum of roots (a+b) = -1/6, product (ab) = -2/6 = -1/3. We need (a/b + b/a) = (a^2 + b^2) / ab = ((a+b)^2 - 2ab) / ab. Substituting: ((-1/6)^2 - 2(-1/3)) / (-1/3) = (1/36 + 2/3) / (-1/3) = (1/36 + 24/36) / (-1/3) = (25/36) * (-3) = -25/12.
Using the relationship between zeros and coefficients for 6x^2 + x - 2, the sum of zeros (a + b) is -1/6 and the product (ab) is -2/6, which simplifies to -1/3. The given expression a/b + b/a can be combined into the single fraction (a^2 + b^2) / ab. By adding and subtracting 2ab in the numerator, we rewrite this as ((a + b)^2 - 2ab) / ab, which is the standard algebraic identity for this form. Substituting the values gives ((-1/6)^2 - 2(-1/3)) / (-1/3), which simplifies to (1/36 + 2/3) / (-1/3) and then to (25/36) / (-1/3). Multiplying the numerator by the reciprocal of the denominator yields 25/36 multiplied by -3, resulting in -25/12.