**Given:**

Solve the following quadratic equations:

I. $$x^2 - 7x - 260 = 0$$

II. $$y^2 - 3y - 54 = 0$$

Based on the solutions, determine the correct relationship between *x* and *y*.

Choose the correct option:

Question

**Given:**

Solve the following quadratic equations:

I. $$x^2 - 7x - 260 = 0$$

II. $$y^2 - 3y - 54 = 0$$

Based on the solutions, determine the correct relationship between *x* and *y*.

Choose the correct option:

Accepted Answer

**Solution:** 1. Solve equation I: $$x^2 - 7x - 260 = 0$$ Using the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-7$, $c=-260$: $$\Delta = (-7)^2 - 4 imes 1 imes (-260) = 49 + 1040 = 1089$$ $$\sqrt{1089} = 33$$ So, $$x = \frac{7 \pm 33}{2}$$ - For plus sign: $$x = \frac{7 + 33}{2} = \frac{40}{2} = 20$$ - For minus sign: $$x = \frac{7 - 33}{2} = \frac{-26}{2} = -13$$ Hence, $$x = 20$$ or $$x = -13$$. 2. Solve equation II: $$y^2 - 3y - 54 = 0$$ Here, $a=1$, $b=-3$, $c=-54$: $$\Delta = (-3)^2 - 4 imes 1 imes (-54) = 9 + 216 = 225$$ $$\sqrt{225} = 15$$ So, $$y = \frac{3 \pm 15}{2}$$ - For plus sign: $$y = \frac{3 + 15}{2} = \frac{18}{2} = 9$$ - For minus sign: $$y = \frac{3 - 15}{2} = \frac{-12}{2} = -6$$ Hence, $$y = 9$$ or $$y = -6$$. 3. Now compare possible values of *x* and *y*: - If $$x = 20$$ and $$y = 9$$, then $$x > y$$. - If $$x = 20$$ and $$y = -6$$, then $$x > y$$. - If $$x = -13$$ and $$y = 9$$, then $$x < y$$. - If $$x = -13$$ and $$y = -6$$, then $$x < y$$. 4. Since *x* can be greater than or less than *y* depending on the roots chosen, **no definite relation** can be established between *x* and *y*. **Answer:** $x = y$ or no relation can be established between $x$ and $y$.

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