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

(I) $ $2x^2 + 7x - 60 = 0 $$

(II) $ $3y^2 - 28y + 64 = 0 $$

Find the correct relation:

Question

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

(I) $ $2x^2 + 7x - 60 = 0 $$

(II) $ $3y^2 - 28y + 64 = 0 $$

Find the correct relation:

Accepted Answer

**Solution:**

1. Solve equation (I):

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

Using the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$, where $a=2$, $b=7$, and $c=-60$:

$$
$\Delta = b^2 - 4ac = 7^2 - 4 \times 2 \times (-60) = 49 + 480 = 529$
$$

$$
$\sqrt{529} = 23$
$$

So,

$$
$x = \frac{-7 \pm 23}{2 \times 2} = \frac{-7 \pm 23}{4}$
$$

Two roots:

- $$ x_1 = \frac{-7 + 23}{4} = \frac{16}{4} = 4$ $
- $ $x_2 = \frac{-7 - 23}{4} = \frac{-30}{4} = -7.5 $$

2. Solve equation (II):

$$ $3y^2 - 28y + 64 = 0$ $$

Here, $a=3$, $b=-28$, $c=64$.

Calculate discriminant:

$$
$\Delta = (-28)^2 - 4 \times 3 \times 64 = 784 - 768 = 16$
$$

$$
$\sqrt{16} = 4$
$$

Roots:

$$
$y = \frac{28 \pm 4}{2 \times 3} = \frac{28 \pm 4}{6}$
$$

- $$ y_1 = \frac{28 + 4}{6} = \frac{32}{6} = \frac{16}{3} \approx 5.33 $$
- $ $y_2 = \frac{28 - 4}{6} = \frac{24}{6} = 4$ $

3. Compare roots:

Possible values of $x$: 4 and -7.5

Possible values of $y$: 5.33 and 4

Check all combinations:

- $x=4$ and $y=5.33$ : $4 < 5.33$
- $x=4$ and $y=4$ : $4 = 4$
- $x=-7.5$ and $y=5.33$ : $-7.5 < 5.33$
- $x=-7.5$ and $y=4$ : $-7.5 < 4$

In all cases, $$x \leq y$$.

**Answer:** $ $x \leq y$ $ (option d)

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