Evaluating $(\sqrt{5} + \sqrt{7})^2$

To determine the nature of the number $(\sqrt{5} + \sqrt{7})^2$, we first need to expand the expression.

We use the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$. In this case, $a = \sqrt{5}$ and $b = \sqrt{7}$.

Expanding the Expression

• Substitute the values into the identity: $(\sqrt{5} + \sqrt{7})^2 = (\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{7}) + (\sqrt{7})^2$
• Calculate the square terms: $(\sqrt{5})^2 = 5$ $(\sqrt{7})^2 = 7$
• Calculate the middle term: $2(\sqrt{5})(\sqrt{7}) = 2\sqrt{5 \times 7} = 2\sqrt{35}$
• Combine the terms: $( \sqrt{5} + \sqrt{7} )^2 = 5 + 2\sqrt{35} + 7$ $( \sqrt{5} + \sqrt{7} )^2 = 12 + 2\sqrt{35}$

Classifying the Result

The expanded expression is $12 + 2\sqrt{35}$. Now, we need to classify this number:

• A rational number can be expressed as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
• An irrational number cannot be expressed as such a fraction.
• $\sqrt{35}$ is an irrational number because 35 is not a perfect square.
• Multiplying an irrational number ($ \sqrt{35} $) by a non-zero integer (2) results in an irrational number ($ 2\sqrt{35} $).
• Adding an integer (12) to an irrational number ($ 2\sqrt{35} $) results in an irrational number.

Therefore, $(\sqrt{5} + \sqrt{7})^2$ is an irrational number.