Question

1. \\(\sqrt{5} \cdot \sqrt{5}\\)
2. \\(\sqrt{12}\\)

Explanation:

Problem 17: $\boldsymbol{\sqrt{5} \cdot \sqrt{5}}$

Step 1: Recall square root multiplication rule

The rule for multiplying square roots is $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$ (for $a \geq 0, b \geq 0$). Here, $a = b = 5$, so we can apply this rule: $\sqrt{5} \cdot \sqrt{5} = \sqrt{5 \cdot 5}$.

Step 2: Simplify the radicand

Simplify the expression inside the square root: $\sqrt{5 \cdot 5} = \sqrt{25}$.

Step 3: Evaluate the square root

The square root of 25 is 5, since $5^2 = 25$. So $\sqrt{25} = 5$.

Step 1: Factor the radicand

Factor 12 into a product of a perfect square and another number. We know that $12 = 4 \cdot 3$, where 4 is a perfect square ($2^2 = 4$).

Step 2: Apply square root property

Using the property $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$ (for $a \geq 0, b \geq 0$), we can rewrite $\sqrt{12}$ as $\sqrt{4 \cdot 3} = \sqrt{4} \cdot \sqrt{3}$.

Step 3: Simplify the perfect square root

Simplify $\sqrt{4}$, which is 2 (since $2^2 = 4$). So $\sqrt{4} \cdot \sqrt{3} = 2\sqrt{3}$.

Answer:

$5$

Problem 20: $\boldsymbol{\sqrt{12}}$