Given the two quadratic equations:

(I) $4x^{2} + 3x - 10 = 0$

(II) $y^{2} + 10y - 119 = 0$

Solve both equations and determine the correct relationship between $x$ and $y$ from the following options.

Question

Given the two quadratic equations:

(I) $4x^{2} + 3x - 10 = 0$

(II) $y^{2} + 10y - 119 = 0$

Solve both equations and determine the correct relationship between $x$ and $y$ from the following options.

Accepted Answer

**Solution:**

1. Solve equation (I): $4x^{2} + 3x - 10 = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ where $a=4$, $b=3$, $c=-10$,

$$
\Delta = b^{2} - 4ac = 3^{2} - 4 	imes 4 	imes (-10) = 9 + 160 = 169
$$

$$
x = \frac{-3 \pm \sqrt{169}}{2 	imes 4} = \frac{-3 \pm 13}{8}
$$

So,
- $x_1 = \frac{-3 + 13}{8} = \frac{10}{8} = 1.25$
- $x_2 = \frac{-3 - 13}{8} = \frac{-16}{8} = -2$

2. Solve equation (II): $y^{2} + 10y - 119 = 0$

Here, $a=1$, $b=10$, $c=-119$,

$$
\Delta = 10^{2} - 4 	imes 1 	imes (-119) = 100 + 476 = 576
$$

$$
y = \frac{-10 \pm \sqrt{576}}{2 	imes 1} = \frac{-10 \pm 24}{2}
$$

So,
- $y_1 = \frac{-10 + 24}{2} = \frac{14}{2} = 7$
- $y_2 = \frac{-10 - 24}{2} = \frac{-34}{2} = -17$

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

- $x$ roots: $1.25$, $-2$
- $y$ roots: $7$, $-17$

Since the roots are distinct and no consistent inequality holds for all pairs (e.g., $x_1 = 1.25 < y_1 = 7$ but $x_2 = -2 > y_2 = -17$), no definite relation between $x$ and $y$ can be established.

**Answer:** If $x = y$ or no relation can be established between $x$ and $y$.

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