Given the two quadratic equations:

I. $2x^2 - x - 3 = 0$

II. $2y^2 - 6y + 4 = 0$

Determine the correct relationship between the roots $x$ and $y$:

Question

Given the two quadratic equations:

I. $2x^2 - x - 3 = 0$

II. $2y^2 - 6y + 4 = 0$

Determine the correct relationship between the roots $x$ and $y$:

Accepted Answer

**Solution:**

1. Solve the first equation:

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

Factorizing:

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

Group terms:

$$2x(x + 1) - 3(x + 1) = 0$$

Take common factor:

$$(x + 1)(2x - 3) = 0$$

So, the roots are:

$$x = -1 \quad 	ext{or} \quad x = \frac{3}{2}$$

2. Solve the second equation:

$$2y^2 - 6y + 4 = 0$$

Divide entire equation by 2:

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

Factorize:

$$(y - 1)(y - 2) = 0$$

So, the roots are:

$$y = 1 \quad 	ext{or} \quad y = 2$$

3. Comparing the roots:

- $x$ can be $-1$ or $1.5$
- $y$ can be $1$ or $2$

Since $x$ and $y$ can take multiple values and there is no consistent inequality or equality between all roots, the correct option is that $x = y$ or no definite relation can be established.

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

Answer

Answer

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