Given the quadratic equations:

I. $(2x^2 + 13x - 150 = 0)$

II. $(y^2 - 17y + 66 = 0)$

Determine the correct relationship between $x$ and $y$:

Question

Given the quadratic equations:

I. $(2x^2 + 13x - 150 = 0)$

II. $(y^2 - 17y + 66 = 0)$

Determine the correct relationship between $x$ and $y$:

Accepted Answer

**Solution:**

1. Consider the first equation:

$$2x^2 + 13x - 150 = 0$$

Factor by grouping:

$$2x^2 - 12x + 25x - 150 = 0$$

$$2x(x - 6) + 25(x - 6) = 0$$

$$(2x + 25)(x - 6) = 0$$

So, the roots are:

$$x = -\frac{25}{2} = -12.5, \quad x = 6$$

2. Now, consider the second equation:

$$y^2 - 17y + 66 = 0$$

Factor by grouping:

$$y^2 - 11y - 6y + 66 = 0$$

$$y(y - 11) - 6(y - 11) = 0$$

$$(y - 6)(y - 11) = 0$$

So, the roots are:

$$y = 6, \quad y = 11$$

3. Comparing the roots:

- For $x$: $-12.5, 6$
   - For $y$: $6, 11$

The maximum root of $x$ is $6$, and the minimum root of $y$ is $6$.

Since $x$ can be $-12.5$ or $6$, and $y$ can be $6$ or $11$, the relation $x \leq y$ holds for all roots.

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

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