Verified Answer :
- Identify the common radical parts in each expression to apply the distributive property.
- Add the coefficients for the first expression: \(2\sqrt{3} + 5\sqrt{3} = (2+5)\sqrt{3} = 7\sqrt{3}\).
- Subtract the coefficients for the second expression, noting \(\sqrt{5} = 1\sqrt{5}\): \(7\sqrt{5} - 1\sqrt{5} = (7-1)\sqrt{5} = 6\sqrt{5}\).
- Final results: \(\boxed{7\sqrt{3}}\) and \(\boxed{6\sqrt{5}}\).
Explanation
- Understanding the Problem and Identifying Like Radicals
We are given two expressions involving radical terms: \(2\sqrt{3} + 5\sqrt{3}\) and \(7\sqrt{5} - \sqrt{5}\). Both expressions consist of what are known as "like radicals," meaning that the radical parts—the roots—are identical in each expression. For the first expression, both terms involve \(\sqrt{3}\). For the second expression, both terms involve \(\sqrt{5}\). Combining like radicals follows the same distributive property used for combining like terms in algebra, such as \(ax + bx = (a+b)x\). In this case, the property is \(a\sqrt{n} + b\sqrt{n} = (a+b)\sqrt{n}\) and \(a\sqrt{n} - b\sqrt{n} = (a-b)\sqrt{n}\).
- Simplifying the First Expression
To simplify \(2\sqrt{3} + 5\sqrt{3}\), we notice that the common radical part is \(\sqrt{3}\). According to the distributive property, we can factor out \(\sqrt{3}\) from both terms:
\[2\sqrt{3} + 5\sqrt{3} = (2 + 5)\sqrt{3}\]
Adding the coefficients \(2\) and \(5\), we get \(7\). Therefore, the expression simplifies to \(7\sqrt{3}\).
- Simplifying the Second Expression
To simplify \(7\sqrt{5} - \sqrt{5}\), we first note that the second term, \(\sqrt{5}\), is equivalent to \(1\sqrt{5}\). The common radical part is \(\sqrt{5}\). Again, using the distributive property:
\[7\sqrt{5} - 1\sqrt{5} = (7 - 1)\sqrt{5}\]
Subtracting the coefficient \(1\) from \(7\), we get \(6\). Therefore, the expression simplifies to \(6\sqrt{5}\).
- Conclusion and Final Results
In summary, we have simplified the two expressions:
The first expression \(2\sqrt{3} + 5\sqrt{3}\) simplifies to \(7\sqrt{3}\).
The second expression \(7\sqrt{5} - \sqrt{5}\) simplifies to \(6\sqrt{5}\).
The final answers are \(\boxed{7\sqrt{3}}\) and \(\boxed{6\sqrt{5}}\).
Examples
Radicals appear in various real-world situations, such as architecture and engineering when calculating distances or dimensions of structures. For example, if you are designing a square garden with a side length of \(\sqrt{3}\) meters and you need to combine two sections of fencing that are \(2\sqrt{3}\) meters and \(5\sqrt{3}\) meters long, you would add the lengths to get a total of \(7\sqrt{3}\) meters. Mastering the addition and subtraction of like radicals is a foundational skill that helps in simplifying complex measurements into manageable figures.