Given the two quadratic equations:

I. $$x^2 - 7x - 60 = 0$$

II. $$y^2 + 13y + 40 = 0$$

Find the correct relationship between the roots *x* and *y* from the options below.

Question

Given the two quadratic equations:

I. $$x^2 - 7x - 60 = 0$$

II. $$y^2 + 13y + 40 = 0$$

Find the correct relationship between the roots *x* and *y* from the options below.

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:
   $$x^2 - 7x - 60 = 0$$

Using the quadratic formula:
   $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
   Here, $a=1$, $b=-7$, $c=-60$.

Calculate the discriminant:
   $$\Delta = (-7)^2 - 4 	imes 1 	imes (-60) = 49 + 240 = 289$$

So,
   $$x = \frac{7 \pm \sqrt{289}}{2} = \frac{7 \pm 17}{2}$$

Roots:
   - $$x_1 = \frac{7 + 17}{2} = \frac{24}{2} = 12$$
   - $$x_2 = \frac{7 - 17}{2} = \frac{-10}{2} = -5$$

2. Solve the second quadratic equation:
   $$y^2 + 13y + 40 = 0$$

Here, $a=1$, $b=13$, $c=40$.

Calculate the discriminant:
   $$\Delta = 13^2 - 4 	imes 1 	imes 40 = 169 - 160 = 9$$

So,
   $$y = \frac{-13 \pm \sqrt{9}}{2} = \frac{-13 \pm 3}{2}$$

Roots:
   - $$y_1 = \frac{-13 + 3}{2} = \frac{-10}{2} = -5$$
   - $$y_2 = \frac{-13 - 3}{2} = \frac{-16}{2} = -8$$

3. Compare the roots:
   - Roots of *x* are 12 and -5
   - Roots of *y* are -5 and -8

The greatest root of *x* is 12, which is greater than the greatest root of *y* (which is -5).

**Answer:** $x > y$

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