Question

Create and solve two problems involving addition, subtraction, multiplication, and division of radicals. The two problems combined should contain all four operations. For example, one problem could combine addition and multiplication and the second problem could contain subtraction and division. Check out the following sample problem to help you get started.

Answer 1

Problem 1:
Solve the following expression: $\sqrt{16} + \sqrt{9} \cdot \sqrt{4} - \sqrt{25} \div \sqrt{5}$.

Solution:
First, let's simplify each radical term individually:
$\sqrt{16} = 4$, since 4 squared is 16,
$\sqrt{9} \cdot \sqrt{4} = 3 \cdot 2 = 6$, and
$\sqrt{25} \div \sqrt{5} = 5 \div \sqrt{5}$.

Now, we can simplify the expression further:
$4 + 6 - 5 \div \sqrt{5}$,
$4 + 6 - \frac{5}{\sqrt{5}}$.

Next, rationalize the denominator of the fraction by multiplying the numerator and denominator by $\sqrt{5}$:
$4 + 6 - \frac{5 \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}}$,
$4 + 6 - \frac{5 \cdot \sqrt{5}}{5}$.

Simplify the expression:
$4 + 6 - \sqrt{5}$,
$10 - \sqrt{5}$.

Therefore, the solution to the expression $\sqrt{16} + \sqrt{9} \cdot \sqrt{4} - \sqrt{25} \div \sqrt{5}$ is $10 - \sqrt{5}$.

Problem 2:
Solve the following equation: $2 \cdot \sqrt{9} - (\sqrt{25} - 1) \div \sqrt{2}$.

Solution:
First, simplify each radical term individually:
$\sqrt{9} = 3$, since 3 squared is 9,
$\sqrt{25} = 5$, and
$\sqrt{2}$ remains the same.

Now, we can solve the equation further:
$2 \cdot 3 - (5 - 1) \div \sqrt{2}$,
$6 - 4 \div \sqrt{2}$.

Next, divide 4 by $\sqrt{2}$:
$6 - \frac{4}{\sqrt{2}}$,
$6 - \frac{4 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}$,
$6 - \frac{4 \cdot \sqrt{2}}{2}$,
$6 - 2 \cdot \sqrt{2}$.

Therefore, the solution to the equation $2 \cdot \sqrt{9} - (\sqrt{25} - 1) \div \sqrt{2}$ is $6 - 2 \cdot \sqrt{2}$.