What is the following product?
$(\sqrt{12}+\sqrt{6})(\sqrt{6}-\sqrt{10})$
Answer (2)
The product ( 12 + 6 ) ( 6 − 10 ) simplifies to 6 2 − 2 30 + 6 − 2 15 . Each step involves expanding the expression using the distributive property and simplifying the resulting terms. The final result cannot be simplified further as there are no like terms.
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Expand the product using the distributive property.
Simplify each term by multiplying the square roots.
Simplify the square roots by finding perfect square factors.
The final simplified expression is 6 2 − 2 30 + 6 − 2 15 .
Explanation
Understanding the Problem We are asked to compute the product of two expressions: ( 12 + 6 ) ( 6 − 10 ) . The expression involves square roots of integers. We need to simplify the expression after expanding the product.
Expanding the Product We will expand the product using the distributive property (also known as the FOIL method). This means we multiply each term in the first parentheses by each term in the second parentheses: ( 12 + 6 ) ( 6 − 10 ) = 12 ⋅ 6 − 12 ⋅ 10 + 6 ⋅ 6 − 6 ⋅ 10
Simplifying Each Term Now, let's simplify each term: 12 ⋅ 6 = 12 × 6 = 72 12 ⋅ 10 = 12 × 10 = 120 6 ⋅ 6 = 6 6 ⋅ 10 = 6 × 10 = 60 So the expression becomes: 72 − 120 + 6 − 60
Simplifying the Square Roots Next, we simplify the square roots by finding the largest perfect square factors: 72 = 36 × 2 = 36 ⋅ 2 = 6 2 120 = 4 × 30 = 4 ⋅ 30 = 2 30 60 = 4 × 15 = 4 ⋅ 15 = 2 15 Substituting these back into the expression, we get: 6 2 − 2 30 + 6 − 2 15
Final Simplified Expression The simplified expression is 6 2 − 2 30 + 6 − 2 15 . There are no like terms to combine, so this is our final answer.
Conclusion Therefore, the product ( 12 + 6 ) ( 6 − 10 ) simplifies to 6 2 − 2 30 + 6 − 2 15 .
Examples
Understanding how to simplify expressions with square roots is very useful in various fields, such as physics and engineering, when dealing with distances, areas, or volumes. For instance, when calculating the length of the diagonal of a rectangle with sides 12 and 6 , you would use the Pythagorean theorem, which involves square roots. Simplifying such expressions allows for more accurate and manageable calculations in real-world applications.
