Simplify the radicals.
[tex]\begin{array}{l}
\sqrt{10} \cdot \sqrt{8}= \\
\sqrt{2} \cdot \sqrt{10}=
\end{array}[/tex]
Answer (2)
Multiply the radicals: 10 ⋅ 8 = 80 and 2 ⋅ 10 = 20 .
Simplify 80 : 80 = 16 × 5 = 4 5 .
Simplify 20 : 20 = 4 × 5 = 2 5 .
The simplified radicals are: 10 ⋅ 8 = 4 5 and 2 ⋅ 10 = 2 5 .
Explanation
Understanding the problem We are asked to simplify two radical expressions: 10 ⋅ 8 and 2 ⋅ 10 . We need to multiply the radicals and then simplify the resulting radical expressions by finding perfect square factors.
Simplifying the first radical First, let's simplify 10 ⋅ 8 .
10 ⋅ 8 = 10 × 8 = 80 Now, we simplify 80 by finding the largest perfect square that divides 80. We know that 80 = 16 × 5 , and 16 is a perfect square ( 16 = 4 2 ). Therefore, 80 = 16 × 5 = 16 ⋅ 5 = 4 5
Simplifying the second radical Next, let's simplify 2 ⋅ 10 .
2 ⋅ 10 = 2 × 10 = 20 Now, we simplify 20 by finding the largest perfect square that divides 20. We know that 20 = 4 × 5 , and 4 is a perfect square ( 4 = 2 2 ). Therefore, 20 = 4 × 5 = 4 ⋅ 5 = 2 5
Final Answer So, we have simplified the two radical expressions: 10 ⋅ 8 = 4 5 2 ⋅ 10 = 2 5
Examples
Radical simplification is useful in various fields, such as engineering and physics, when dealing with lengths, areas, or volumes that involve square roots. For example, when calculating the length of the diagonal of a square with side length s , the diagonal is s 2 . Simplifying radicals helps in obtaining the most concise and understandable form of such measurements. Also, in circuit analysis, simplifying radicals can help in determining impedance or resonant frequency in AC circuits, making calculations easier and more intuitive.
To simplify 10 ⋅ 8 we get 4 5 , and for 2 ⋅ 10 we obtain 2 5 . The process involved applying the product property of square roots and then simplifying the resulting radicals by factoring. Both results arrive at a more concise radical form.
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