Given the two quadratic equations:

I. $$x^2 + x - 12 = 0$$

II. $$y^2 + 2y - 15 = 0$$

Solve both equations and determine the correct relationship between *x* and *y* from the following options.

Question

Given the two quadratic equations:

I. $$x^2 + x - 12 = 0$$

II. $$y^2 + 2y - 15 = 0$$

Solve both equations and determine the correct relationship between *x* and *y* from the following options.

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:
   $$x^2 + x - 12 = 0$$

Using the quadratic formula:
   $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
   where $a=1$, $b=1$, and $c=-12$.

Calculate the discriminant:
   $$\Delta = 1^2 - 4 	imes 1 	imes (-12) = 1 + 48 = 49$$

So,
   $$x = \frac{-1 \pm \sqrt{49}}{2} = \frac{-1 \pm 7}{2}$$

Therefore,
   - $$x_1 = \frac{-1 + 7}{2} = \frac{6}{2} = 3$$
   - $$x_2 = \frac{-1 - 7}{2} = \frac{-8}{2} = -4$$

2. Solve the second quadratic equation:
   $$y^2 + 2y - 15 = 0$$

Here, $a=1$, $b=2$, and $c=-15$.

Calculate the discriminant:
   $$\Delta = 2^2 - 4 	imes 1 	imes (-15) = 4 + 60 = 64$$

So,
   $$y = \frac{-2 \pm \sqrt{64}}{2} = \frac{-2 \pm 8}{2}$$

Therefore,
   - $$y_1 = \frac{-2 + 8}{2} = \frac{6}{2} = 3$$
   - $$y_2 = \frac{-2 - 8}{2} = \frac{-10}{2} = -5$$

3. Now, compare the roots of *x* and *y*:
   - Possible values of *x*: 3, -4
   - Possible values of *y*: 3, -5

4. Check all possible comparisons:
   - $x=3$ and $y=3$ → $x = y$
   - $x=3$ and $y=-5$ → $x > y$
   - $x=-4$ and $y=3$ → $x < y$
   - $x=-4$ and $y=-5$ → $x > y$

5. Since the roots can have different relations depending on which roots are chosen, but the question implies comparing the roots generally, the largest root of *x* (3) is greater than the largest root of *y* (3), they are equal in that case.

6. However, the question likely expects the comparison of the greater roots, which are $x=3$ and $y=3$, so $x = y$.

**Answer:** (e)

Answer

Answer

Answer

Answer