Given the two quadratic equations:

I. $$x^2 - 25x + 156 = 0$$

II. $$y^2 - 27y + 182 = 0$$

If $$x$$ and $$y$$ are the roots of the respective equations, determine the correct relationship between $$x$$ and $$y$$.

Question

Given the two quadratic equations:

I. $$x^2 - 25x + 156 = 0$$

II. $$y^2 - 27y + 182 = 0$$

If $$x$$ and $$y$$ are the roots of the respective equations, determine the correct relationship between $$x$$ and $$y$$.

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:
   $$x^2 - 25x + 156 = 0$$

Using the quadratic formula:
   $$x = \frac{25 \pm \sqrt{25^2 - 4 	imes 1 	imes 156}}{2} = \frac{25 \pm \sqrt{625 - 624}}{2} = \frac{25 \pm 1}{2}$$

So, the roots are:
   - $$x_1 = \frac{25 + 1}{2} = 13$$
   - $$x_2 = \frac{25 - 1}{2} = 12$$

2. Solve the second quadratic equation:
   $$y^2 - 27y + 182 = 0$$

Using the quadratic formula:
   $$y = \frac{27 \pm \sqrt{27^2 - 4 	imes 1 	imes 182}}{2} = \frac{27 \pm \sqrt{729 - 728}}{2} = \frac{27 \pm 1}{2}$$

So, the roots are:
   - $$y_1 = \frac{27 + 1}{2} = 14$$
   - $$y_2 = \frac{27 - 1}{2} = 13$$

3. Compare the roots:
   - The roots of equation I are 12 and 13.
   - The roots of equation II are 13 and 14.

Since $$x$$ can be either 12 or 13, and $$y$$ can be either 13 or 14, the smallest $$x$$ is 12 and the smallest $$y$$ is 13.

Therefore, for all possible values, $$x \leq y$$ holds true.

**Answer:** $$x \leq y$$

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