In each question, two equations numbered (I) and (II) are given. Solve both the equations and determine the correct relationship between *x* and *y*.

(I) $ x^2 + 14x + 45 = 0 $

(II) \( 3y^2 - y - 10 = 0:

Question

In each question, two equations numbered (I) and (II) are given. Solve both the equations and determine the correct relationship between *x* and *y*.

(I) $ x^2 + 14x + 45 = 0 $

(II) \( 3y^2 - y - 10 = 0:

Accepted Answer

**Solution:**

1. Solve equation (I):

$$ x^2 + 14x + 45 = 0 $$

Calculate the discriminant:

$$ D = $14^2 - 4 \times 1 \times 45 = 196 - 180 = 16$ $$

Roots of the quadratic are:

$$ x = $\frac{-14 \pm \sqrt{16}}{2} = \frac{-14 \pm 4}{2}$ $$

So,

- $ x_1 = $\frac{-14 + 4}{2} = \frac{-10}{2}$ = -5 $
- $ x_2 = $\frac{-14 - 4}{2} = \frac{-18}{2}$ = -9 $

2. Solve equation (II):

$$ 3y^2 - y - 10 = 0 $$

Calculate the discriminant:

$$ $D = (-1)^2 - 4 \times 3 \times (-10) = 1 + 120 = 121$ $$

Roots of the quadratic are:

$$ y = $\frac{1 \pm \sqrt{121}}{2 \times 3} = \frac{1 \pm 11}{6}$ $$

So,

- $ y_1 = $\frac{1 + 11}{6} = \frac{12}{6}$ = 2 $
- $ y_2 = $\frac{1 - 11}{6} = \frac{-10}{6}$ = -$\frac{5}{3} \approx -1.67$ $

3. Compare the roots:

- Possible values of $ x $ are $ -5 $ and $ -9 $
- Possible values of $ y $ are $ 2 $ and $ -1.67 $

Since both $ x $ values are less than both $ y $ values, it follows that $ x < y $.

**Answer:** (c) $ x < y $

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