Given the two quadratic equations:

I. $(x^2 - 3x - 40 = 0)$

II. $(y^2 + 11y + 30 = 0)$

Which of the following relations between $x$ and $y$ is correct?

Question

Given the two quadratic equations:

I. $(x^2 - 3x - 40 = 0)$

II. $(y^2 + 11y + 30 = 0)$

Which of the following relations between $x$ and $y$ is correct?

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:

$$x^2 - 3x - 40 = 0$$

Factorizing:

$$x^2 - 8x + 5x - 40 = 0$$

$$x(x - 8) + 5(x - 8) = 0$$

$$(x - 8)(x + 5) = 0$$

So, $x = 8$ or $x = -5$.

2. Solve the second quadratic equation:

$$y^2 + 11y + 30 = 0$$

Factorizing:

$$y^2 + 6y + 5y + 30 = 0$$

$$y(y + 6) + 5(y + 6) = 0$$

$$(y + 5)(y + 6) = 0$$

So, $y = -5$ or $y = -6$.

3. Compare the roots:

- Possible values of $x$ are 8 and -5.
   - Possible values of $y$ are -5 and -6.

The maximum value of $x$ is 8, and the maximum value of $y$ is -5.

Since $8 \geq -5$ and $-5 \geq -6$, it follows that $x \geq y$.

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

Answer

Answer

Answer

Answer