Given the two quadratic equations:

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

II. $$y^2 + 11y + 30 = 0$$

If $$x$$ and $$y$$ satisfy these equations respectively, determine the correct relationship between $$x$$ and $$y$$ from the following options.

Question

Given the two quadratic equations:

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

II. $$y^2 + 11y + 30 = 0$$

If $$x$$ and $$y$$ satisfy these equations respectively, determine the correct relationship between $$x$$ and $$y$$ from the following options.

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:

$$x^2 + x - 20 = 0$$

Using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Here, $$a=1$$, $$b=1$$, and $$c=-20$$.

Calculate the discriminant:

$$\Delta = 1^2 - 4 	imes 1 	imes (-20) = 1 + 80 = 81$$

So,

$$x = \frac{-1 \pm \sqrt{81}}{2} = \frac{-1 \pm 9}{2}$$

Therefore,

- $$x_1 = \frac{-1 + 9}{2} = \frac{8}{2} = 4$$

- $$x_2 = \frac{-1 - 9}{2} = \frac{-10}{2} = -5$$

2. Solve the second quadratic equation:

$$y^2 + 11y + 30 = 0$$

Here, $$a=1$$, $$b=11$$, and $$c=30$$.

Calculate the discriminant:

$$\Delta = 11^2 - 4 	imes 1 	imes 30 = 121 - 120 = 1$$

So,

$$y = \frac{-11 \pm \sqrt{1}}{2} = \frac{-11 \pm 1}{2}$$

Therefore,

- $$y_1 = \frac{-11 + 1}{2} = \frac{-10}{2} = -5$$

- $$y_2 = \frac{-11 - 1}{2} = \frac{-12}{2} = -6$$

3. Now, compare the roots of $$x$$ and $$y$$:

- $$x$$ roots: 4 and -5

- $$y$$ roots: -5 and -6

4. Since $$x$$ and $$y$$ are roots of their respective equations, the possible values are:

- $$x = 4$$ or $$-5$$

- $$y = -5$$ or $$-6$$

5. Check all combinations:

- If $$x = 4$$ and $$y = -6$$, then $$x > y$$

- If $$x = 4$$ and $$y = -5$$, then $$x > y$$

- If $$x = -5$$ and $$y = -6$$, then $$x > y$$

- If $$x = -5$$ and $$y = -5$$, then $$x = y$$

6. From these, the relationship $$x \geq y$$ always holds true.

**Answer:** If $$x \geq y$$

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