Given that $ x - rac{3}{2} = rac{1}{2} imes rac{5 ext{√}128}{ ext{√}32} $, find the value of $ x $, where $ x ext{ } ext{ } ext{ } ext{} ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } \ \ A = \{\text{√}32k^{2}, \text{√}50k^{2}, \text{√}128k^{2}, \text{√}98k^{2} \} , where \ k ext{ } \in \mathbb{N} - Leaving Cert Mathematics - Question 6 - 2022

To find the values in set A:

• $A = \{\text{√}32k^{2}, \text{√}50k^{2}, \text{√}128k^{2}, \text{√}98k^{2}\}$

Calculating each term:

• $\text{√}32k^{2} = 4k\text{√}2$
• $\text{√}50k^{2} = 5k\text{√}2$
• $\text{√}128k^{2} = 8k\text{√}2$
• $\text{√}98k^{2} = 7k\text{√}2$

Arrange the terms in ascending order:

• $4k\text{√}2, 5k\text{√}2, 7k\text{√}2, 8k\text{√}2$

Mean Calculation: $\text{Mean} = \frac{(4k\text{√}2 + 5k\text{√}2 + 7k\text{√}2 + 8k\text{√}2)}{4} = \frac{24k\text{√}2}{4} = 6k\text{√}2$

Median Calculation: For a set of four numbers, the median is given by: $\text{Median} = \frac{(5k\text{√}2 + 7k\text{√}2)}{2} = \frac{12k\text{√}2}{2} = 6k\text{√}2$

Thus, the mean is equal to the median.

Assume that $\text{√}2$ is rational.

Then, we can express it as: $\text{√}2 = \frac{p}{q}$ where $p$ and $q$ are integers with no common factors.

Squaring both sides gives: $2 = \frac{p^{2}}{q^{2}} \Rightarrow p^{2} = 2q^{2}$

From this, we conclude that:

1. $p^{2}$ is even, so $p$ must also be even.
2. Let $p = 2k$ for some integer $k$.
3. Substituting back gives: $(2k)^{2} = 2q^{2} \Rightarrow 4k^{2} = 2q^{2} \Rightarrow q^{2} = 2k^{2}$

This implies that:

1. $q^{2}$ is also even, hence $q$ must be even.

Since both p and q are even, this contradicts our assertion that they had no common factors. Therefore,: $\text{√}2 cannot be rational.$