Given the two quadratic equations:

I. $$x^2 - 20x + 96 = 0$$

II. $$y^2 - 3y - 28 = 0$$

Find the correct relationship between *x* and *y*.

Question

Given the two quadratic equations:

I. $$x^2 - 20x + 96 = 0$$

II. $$y^2 - 3y - 28 = 0$$

Find the correct relationship between *x* and *y*.

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:
   $$x^2 - 20x + 96 = 0$$

Calculate the discriminant:
   $$D_x = (-20)^2 - 4 	imes 1 	imes 96 = 400 - 384 = 16$$

Roots of equation I:
   $$x = \frac{20 \pm \sqrt{16}}{2} = \frac{20 \pm 4}{2}$$

So,
   $$x_1 = \frac{20 + 4}{2} = 12$$
   $$x_2 = \frac{20 - 4}{2} = 8$$

2. Solve the second quadratic equation:
   $$y^2 - 3y - 28 = 0$$

Calculate the discriminant:
   $$D_y = (-3)^2 - 4 	imes 1 	imes (-28) = 9 + 112 = 121$$

Roots of equation II:
   $$y = \frac{3 \pm \sqrt{121}}{2} = \frac{3 \pm 11}{2}$$

So,
   $$y_1 = \frac{3 + 11}{2} = 7$$
   $$y_2 = \frac{3 - 11}{2} = -4$$

3. Now, compare the roots:
   - Possible values of *x* are 12 and 8
   - Possible values of *y* are 7 and -4

4. Since both *x* values (12 and 8) are greater than both *y* values (7 and -4), the relationship between *x* and *y* is:

$$x > y$$

**Answer:** If $$x > y$$

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