Given the two quadratic equations:

I. $$x^2 + 31x + 84 = 0$$

II. $$4y^2 - 19y + 21 = 0$$

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

Question

Given the two quadratic equations:

I. $$x^2 + 31x + 84 = 0$$

II. $$4y^2 - 19y + 21 = 0$$

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

Accepted Answer

**Solution:**

1. **Solve equation I:**

$$x^2 + 31x + 84 = 0$$

Calculate the discriminant:

$$\Delta = 31^2 - 4 	imes 1 	imes 84 = 961 - 336 = 625$$

Roots are:

$$x = \frac{-31 \pm \sqrt{625}}{2} = \frac{-31 \pm 25}{2}$$

So,

- $$x_1 = \frac{-31 + 25}{2} = \frac{-6}{2} = -3$$
- $$x_2 = \frac{-31 - 25}{2} = \frac{-56}{2} = -28$$

2. **Solve equation II:**

$$4y^2 - 19y + 21 = 0$$

Calculate the discriminant:

$$\Delta = (-19)^2 - 4 	imes 4 	imes 21 = 361 - 336 = 25$$

Roots are:

$$y = \frac{19 \pm \sqrt{25}}{2 	imes 4} = \frac{19 \pm 5}{8}$$

So,

- $$y_1 = \frac{19 + 5}{8} = \frac{24}{8} = 3$$
- $$y_2 = \frac{19 - 5}{8} = \frac{14}{8} = 1.75$$

3. **Compare roots:**

- Roots of $$x$$ are $$-3$$ and $$-28$$
- Roots of $$y$$ are $$3$$ and $$1.75$$

Since both $$x$$ roots are negative and both $$y$$ roots are positive, any $$x$$ root is less than any $$y$$ root.

**Answer:** $$x < y$$

Answer

Answer

Answer

Answer