Given the two quadratic equations:

I. $ x^2 - 15x + 44 = 0 $

II. $ y^2 - 11y - 80 = 0 $

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

Question

Given the two quadratic equations:

I. $ x^2 - 15x + 44 = 0 $

II. $ y^2 - 11y - 80 = 0 $

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

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:

$$ x^2 - 15x + 44 = 0 $$

Calculate the discriminant:

$$ D_x = ($-15)^2 - 4 \times 1 \times 44 = 225 - 176 = 49 $$$

Roots are:

$$ $x = \frac{15 \pm \sqrt{49}}{2} = \frac{15 \pm 7}{2}$ $$

So,

- $ $x_1 = \frac{15 + 7}{2} = \frac{22}{2} = 11$ $
- $ $x_2 = \frac{15 - 7}{2} = \frac{8}{2} = 4 $$

2. Solve the second quadratic equation:

$$ y^2 - 11y - 80 = 0 $$

Calculate the discriminant:

$$$ D_y = (-11)^2 - 4 \times 1 \times (-80) = 121 + 320 = 441$ $$

Roots are:

$$ $y = \frac{11 \pm \sqrt{441}}{2} = \frac{11 \pm 21}{2}$ $$

So,

- $ $y_1 = \frac{11 + 21}{2} = \frac{32}{2} = 16$ $
- $ $y_2 = \frac{11 - 21}{2} = \frac{-10}{2} = -54 $

3. Now, the possible values of $x$ are 11 and 4, and for $y$ are 16 and -5.

4. Comparing the roots:

- $x_1 = 11$ is less than $y_1 = 16$
- $x_1 = 11$ is greater than $y_2 = -5$
- $x_2 = 4$ is less than $y_1 = 16$
- $x_2 = 4$ is greater than $y_2 = -5$

Since $x$ and $y$ can take values such that $x > y$ or $x < y$, no definite relationship can be established between $x$ and $y$.

**Answer:** (e) $x = y$ or no relation can be established

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