Given the two quadratic equations:

I. $$2x^2 + x - 28 = 0$$

II. $$2y^2 - 23y + 56 = 0$$

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

Question

Given the two quadratic equations:

I. $$2x^2 + x - 28 = 0$$

II. $$2y^2 - 23y + 56 = 0$$

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

Accepted Answer

**Solution:**

1. Solve the first equation:

$$2x^2 + x - 28 = 0$$

Factorizing,

$$2x^2 + 8x - 7x - 28 = 0$$

$$2x(x + 4) - 7(x + 4) = 0$$

$$(2x - 7)(x + 4) = 0$$

Thus, the roots are:

$$x = \frac{7}{2} = 3.5 \quad \text{or} \quad x = -4$$

2. Solve the second equation:

$$2y^2 - 23y + 56 = 0$$

Factorizing,

$$2y^2 - 16y - 7y + 56 = 0$$

$$2y(y - 8) - 7(y - 8) = 0$$

$$(2y - 7)(y - 8) = 0$$

So, the roots are:

$$y = \frac{7}{2} = 3.5 \quad \text{or} \quad y = 8$$

3. Comparing the roots:

- Possible values of *x* are $-4, 3.5$
   - Possible values of *y* are $3.5, 8$

4. Since the maximum value of *x* is 3.5 and the minimum value of *y* is 3.5, and *y* can also be 8 which is greater than any *x*, it follows that:

$$y \geq x$$

Answer

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