Given the two quadratic equations:

(I) $$ x^2 + x - 56 = 0 $$

(II) $ $y^2 + 17y + 72 = 0$ $

Find the correct relation between $ x $ and $ y $ based on the roots of these equations.

Question

Given the two quadratic equations:

(I) $$ x^2 + x - 56 = 0 $$

(II) $ $y^2 + 17y + 72 = 0$ $

Find the correct relation between $ x $ and $ y $ based on the roots of these equations.

Accepted Answer

**Solution:**

1. Solve equation (I): $ $x^2 + x - 56 = 0$ $

Using the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Here, $ a=1, b=1, c=-56 $

Calculate the discriminant:

$$ \Delta = 1^2 - 4 \times 1 \times (-56) = 1 + 224 = 225 $$

Square root of discriminant:

$$ \sqrt{225} = 15 $$

So,

$$ x = \frac{-1 \pm 15}{2} $$

Two roots:

- $ $x_1 = \frac{-1 + 15}{2} = \frac{14}{2} = 7 $$
- $$ x_2 = \frac{-1 - 15}{2} = \frac{-16}{2} = -8 $$

2. Solve equation (II): $ y^2 + 17y + 72 = 0 $

Here, $ a=1, b=17, c=72 $

Calculate discriminant:

$$ \Delta = 17^2 - 4 \times 1 \times 72 = 289 - 288 = 1 $$

Square root:

$$ \sqrt{1} = 1 $$

Roots:

$$ y = \frac{-17 \pm 1}{2} $$

- $$ y_1 = \frac{-17 + 1}{2} = \frac{-16}{2} = -8 $$
- $$ y_2 = \frac{-17 - 1}{2} = \frac{-18}{2} = -9 $$

3. Compare roots:

- $ x $ roots: 7 and -8
- $ y $ roots: -8 and -9

The largest root of $ x $ is 7, which is greater than the largest root of $ y $ which is -8.

The smallest root of $ x $ is -8, which is greater than the smallest root of $ y $ which is -9.

Hence, for all roots, $ x \geq y $.

**Answer:** (e) $ x \geq y $

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