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 $y$.

Choose the correct option:

(a) $x y$

(b) $x \geq y$

(c) $y x$

(d) $y \geq x$

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

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 $y$.

Choose the correct option:

(a) $x > y$

(b) $x \geq y$

(c) $y > x$

(d) $y \geq x$

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

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:

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

3. Comparing the 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 also have other roots where $x > y$ or $y > x$, no definite relationship can be established between $x$ and $y$.

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

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