Given the two quadratic equations:

I. $(2x^2 + 14x - 16 = 0)$

II. $(y^2 - y - 12 = 0)$

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

Question

Given the two quadratic equations:

I. $(2x^2 + 14x - 16 = 0)$

II. $(y^2 - y - 12 = 0)$

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

Accepted Answer

**Solution:**

1. Solve the first quadratic equation:

$$2x^2 + 14x - 16 = 0$$

Divide the equation into two parts for factorization:

$$2x^2 + 16x - 2x - 16 = 0$$

Group terms:

$$2x(x + 8) - 2(x + 8) = 0$$

Factor out common binomial:

$$(2x - 2)(x + 8) = 0$$

Set each factor to zero:

$$2x - 2 = 0 \Rightarrow x = 1$$

$$x + 8 = 0 \Rightarrow x = -8$$

So, the roots of equation I are $x = 1$ and $x = -8$.

2. Solve the second quadratic equation:

$$y^2 - y - 12 = 0$$

Rewrite for factorization:

$$y^2 - 4y + 3y - 12 = 0$$

Group terms:

$$y(y - 4) + 3(y - 4) = 0$$

Factor common binomial:

$$(y + 3)(y - 4) = 0$$

Set each factor to zero:

$$y + 3 = 0 \Rightarrow y = -3$$

$$y - 4 = 0 \Rightarrow y = 4$$

Thus, the roots of equation II are $y = -3$ and $y = 4$.

3. Compare the roots:

- $x$ can be either $1$ or $-8$
- $y$ can be either $4$ or $-3$

Since $x$ and $y$ can take multiple values, no definite relation such as $x \geq y$ or $x < y$ can be established between them.

**Answer:** (e) $x = y$ or no relation.

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