In each of the following questions, two quadratic equations (I) and (II) are given. Solve both equations and determine the correct relationship between *x* and *y*.

I. $$6x^2 + x - 15 = 0$$

II. $$4y^2 - 24y + 35 = 0$$

Question

In each of the following questions, two quadratic equations (I) and (II) are given. Solve both equations and determine the correct relationship between *x* and *y*.

I. $$6x^2 + x - 15 = 0$$

II. $$4y^2 - 24y + 35 = 0$$

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:

$$6x^2 + x - 15 = 0$$

Using the quadratic formula:
   $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
   where $$a = 6$$, $$b = 1$$, and $$c = -15$$.

Calculate the discriminant:
   $$\Delta = 1^2 - 4 	imes 6 	imes (-15) = 1 + 360 = 361$$

So,
   $$x = \frac{-1 \pm \sqrt{361}}{12} = \frac{-1 \pm 19}{12}$$

Two roots:
   - $$x_1 = \frac{-1 + 19}{12} = \frac{18}{12} = 1.5$$
   - $$x_2 = \frac{-1 - 19}{12} = \frac{-20}{12} = -\frac{5}{3} \approx -1.6667$$

2. Solve the second quadratic equation:

$$4y^2 - 24y + 35 = 0$$

Here, $$a = 4$$, $$b = -24$$, $$c = 35$$.

Calculate the discriminant:
   $$\Delta = (-24)^2 - 4 	imes 4 	imes 35 = 576 - 560 = 16$$

So,
   $$y = \frac{24 \pm \sqrt{16}}{8} = \frac{24 \pm 4}{8}$$

Two roots:
   - $$y_1 = \frac{24 + 4}{8} = \frac{28}{8} = 3.5$$
   - $$y_2 = \frac{24 - 4}{8} = \frac{20}{8} = 2.5$$

3. Now compare the roots:

- For $$x$$: $$1.5$$ and $$-1.6667$$
   - For $$y$$: $$3.5$$ and $$2.5$$

The maximum value of $$x$$ is $$1.5$$, which is less than the minimum value of $$y$$ which is $$2.5$$.

Hence, for all roots, $$x < y$$.

**Answer:** If $$x < y$$

Answer

Answer

Answer

Answer