Given the quadratic equations:

I. $(3x^2 - 37x + 114 = 0)$

II. $(3y^2 - 46y + 171 = 0)$

Determine the correct relationship between $x$ and $y$ where $x$ and $y$ are roots of the respective equations.

Question

Given the quadratic equations:

I. $(3x^2 - 37x + 114 = 0)$

II. $(3y^2 - 46y + 171 = 0)$

Determine the correct relationship between $x$ and $y$ where $x$ and $y$ are roots of the respective equations.

Accepted Answer

**Solution:**

1. Consider the first quadratic equation:
   $$
   3x^2 - 37x + 114 = 0
   $$
   Factorize by splitting the middle term:
   $$
   3x^2 - 19x - 18x + 114 = 0
   $$
   Group terms:
   $$
   x(3x - 19) - 6(3x - 19) = 0
   $$
   Factor out common binomial:
   $$
   (x - 6)(3x - 19) = 0
   $$
   So, the roots are:
   $$
   x = 6 \quad \text{or} \quad x = \frac{19}{3} \approx 6.33
   $$

2. Now, consider the second quadratic equation:
   $$
   3y^2 - 46y + 171 = 0
   $$
   Split the middle term:
   $$
   3y^2 - 27y - 19y + 171 = 0
   $$
   Group terms:
   $$
   3y(y - 9) - 19(y - 9) = 0
   $$
   Factor common binomial:
   $$
   (3y - 19)(y - 9) = 0
   $$
   Hence, the roots are:
   $$
   y = \frac{19}{3} \approx 6.33 \quad \text{or} \quad y = 9
   $$

3. Comparing the roots:
   - The roots of $x$ are $6$ and $6.33$
   - The roots of $y$ are $6.33$ and $9$

4. Since the smallest root of $y$ equals the largest root of $x$, and $y$ also has a larger root $9$, it follows that:
   $$
   x \leq y
   $$

**Answer:** $x \leq y$ (Option **d**)

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