Given the two quadratic equations:

I. $ 2x^2 - 7x - 60 = 0 $

II. $ 3y^2 + 13y + 4 = 0 $

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

Question

Given the two quadratic equations:

I. $ 2x^2 - 7x - 60 = 0 $

II. $ 3y^2 + 13y + 4 = 0 $

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

Accepted Answer

**Solution:**

1. Solve the first equation:

$$
2x^2 - 7x - 60 = 0
$$

Factorizing:

$$
2x^2 - 15x + 8x - 60 = 0
$$
$$
x(2x - 15) + 4(2x - 15) = 0
$$
$$
(x + 4)(2x - 15) = 0
$$

So, the roots are:

$$
x = -4 \quad \text{or} \quad x = \frac{15}{2} = 7.5
$$

2. Solve the second equation:

$$
3y^2 + 13y + 4 = 0
$$

Factorizing:

$$
3y^2 + 12y + y + 4 = 0
$$
$$
3y(y + 4) + 1(y + 4) = 0
$$
$$
(3y + 1)(y + 4) = 0
$$

So, the roots are:

$$
y = -4 \quad \text{or} \quad y = -\frac{1}{3}
$$

3. Comparing roots:

- $ x $ can be $-4$ or $7.5$
- $ y $ can be $-4$ or $-\frac{1}{3}$

Since $ x $ and $ y $ share a common root $-4$, but the other roots differ and no consistent inequality holds for all roots, no definite relation between $ x $ and $ y $ can be established.

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

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