Equation Analysis

We are given the equation $\text{a}^2 + \text{b}^2 = 4(\text{a} + 3\text{b} - 10)$, where $\text{a}$ and $\text{b}$ are real numbers. We need to determine the relationship between $\text{a}$ and $\text{b}$ based on this equation.

Let's rearrange the equation to bring all terms to one side:

$\text{a}^2 + \text{b}^2 = 4\text{a} + 12\text{b} - 40$

$\text{a}^2 - 4\text{a} + \text{b}^2 - 12\text{b} + 40 = 0$

Completing the Square

To simplify this equation and reveal the relationship between $\text{a}$ and $\text{b}$, we can use the technique of completing the square for the terms involving $\text{a}$ and the terms involving $\text{b}$.

For the terms involving $\text{a}$ ($\text{a}^2 - 4\text{a}$), we add and subtract $(\frac{-4}{2})^2 = (-2)^2 = 4$:

$\text{a}^2 - 4\text{a} + 4 - 4 = (\text{a} - 2)^2 - 4$

For the terms involving $\text{b}$ ($\text{b}^2 - 12\text{b}$), we add and subtract $(\frac{-12}{2})^2 = (-6)^2 = 36$:

$\text{b}^2 - 12\text{b} + 36 - 36 = (\text{b} - 6)^2 - 36$

Substitute these back into the rearranged equation:

$[(\text{a} - 2)^2 - 4] + [(\text{b} - 6)^2 - 36] + 40 = 0$

$(\text{a} - 2)^2 - 4 + (\text{b} - 6)^2 - 36 + 40 = 0$

$(\text{a} - 2)^2 + (\text{b} - 6)^2 - 40 + 40 = 0$

$(\text{a} - 2)^2 + (\text{b} - 6)^2 = 0$

Determining Relationship between a and b

We have the equation $(\text{a} - 2)^2 + (\text{b} - 6)^2 = 0$. Since $\text{a}$ and $\text{b}$ are real numbers, their squares $(\text{a} - 2)^2$ and $(\text{b} - 6)^2$ must be non-negative (greater than or equal to zero).

• $(\text{a} - 2)^2 \ge 0$
• $(\text{b} - 6)^2 \ge 0$

The sum of two non-negative numbers is zero if and only if both numbers are zero.

Therefore, for the equation $(\text{a} - 2)^2 + (\text{b} - 6)^2 = 0$ to be true, we must have:

• $(\text{a} - 2)^2 = 0 \implies \text{a} - 2 = 0 \implies \text{a} = 2$
• $(\text{b} - 6)^2 = 0 \implies \text{b} - 6 = 0 \implies \text{b} = 6$

So, the only real values of $\text{a}$ and $\text{b}$ that satisfy the given equation are $\text{a} = 2$ and $\text{b} = 6$.

Now, let's compare the values of $\text{a}$ and $\text{b}$:

$\text{a} = 2$

$\text{b} = 6$

Clearly, $2 < 6$.

Thus, the relationship between $\text{a}$ and $\text{b}$ is $\text{a} < \text{b}$.

This means the statement "$\text{a} < \text{b}$" is true.