Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 8: Problem 174
Simplify. (a) \(\sqrt{72}-\sqrt{98}\) (b) \(\sqrt[3]{24}+\sqrt[3]{81}\)(c) \(\frac{1}{2} \sqrt[4]{80}-\frac{2}{3} \sqrt[4]{405}\)
Short Answer
Expert verified
(a) -\root 2\, (b) 5\root[3]{3}\, (c) -\root[4]{5}\).
Step by step solution
01
Simplify \(\sqrt{72}\)
First, simplify \(\sqrt{72}\).\ Since 72 = 36 \times 2, and 36 is a perfect square \(\sqrt{36} = 6\), we can write \(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\).
02
Simplify \(\sqrt{98}\)
Next, simplify \(\sqrt{98}\). Since 98 = 49 \times 2, and 49 is a perfect square \(\sqrt{49} = 7\), we can write \(\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}\).
03
Subtract Simplified Radicals
Now subtract the simplified radicals: \(\sqrt{72} - \sqrt{98} = 6\sqrt{2} - 7\sqrt{2} = (6 - 7)\sqrt{2} = -\sqrt{2}\).
04
Simplify \(\sqrt[3]{24}\)
First, simplify \(\sqrt[3]{24}\). Since 24 = 8\times 3, and 8 is a perfect cube \(\sqrt[3]{8} = 2\), we can write \(\sqrt[3]{24} = \sqrt[3]{8 \times 3} = 2\root[3]{3}\).
05
Simplify \(\sqrt[3]{81}\)
Next, simplify \(\sqrt[3]{81}\). Since 81 = 27 \times 3, and 27 is a perfect cube \(\sqrt[3]{27} = 3\), we can write \(\sqrt[3]{81} = \sqrt[3]{27\times3} = 3\root[3]{3}\).
06
Add Simplified Cube Roots
Now add the simplified cube roots: \(\sqrt[3]{24} + \sqrt[3]{81} = 2\root[3]{3} + 3\root[3]{3} = (2 + 3)\root[3]{3} = 5\root[3]{3}\).
07
Simplify \(\sqrt[4]{80}\)
First, simplify \(\sqrt[4]{80}\). Since 80 = 16\times5, and 16 is a perfect fourth power \(\sqrt[4]{16} = 2\), we can write \(\sqrt[4]{80} = \sqrt[4]{16 \times 5} = 2\root[4]{5}\).
08
Simplify \(\sqrt[4]{405}\)
Next, simplify \(\sqrt[4]{405}\). Since 405 = 81 \times 5, and 81 is a perfect fourth power \(\sqrt[4]{81} = 3\), we can write \(\sqrt[4]{405} = \sqrt[4]{81\times5} = 3\root[4]{5}\).
09
Compute \(\frac{1}{2} \sqrt[4]{80}\)
Calculate \(\frac{1}{2} \sqrt[4]{80}\ = \frac{1}{2}(2 \root[4]{5}) = \root[4]{5}\).
10
Compute \(\frac{2}{3} \sqrt[4]{405}\)
Calculate \(\frac{2}{3} \sqrt[4]{405}\ = \frac{2}{3}(3 \root[4]{5}) = 2\root[4]{5}\).
11
Subtract Simplified Fourth Roots
Subtract the simplified fourth roots: \(\frac{1}{2} \sqrt[4]{80} - \frac{2}{3} \sqrt[4]{405} = \root[4]{5} - 2\root[4]{5} = (1 - 2)\root[4]{5} = -\root[4]{5}\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
square roots
Square roots are a way to find a number which when multiplied by itself gives the original number. For example, the square root of 9 is 3, because 3 * 3 = 9.
When simplifying square roots, look for perfect squares within the number. In the exercise, \(\root{72}\), we broke it down into \(36 \times 2\) because 36 is a perfect square. Then, we found \(\root{36} = 6\), so \(\root{72} = 6\root{2}\).
Similarly, for \(\root{98}\), we used that 98 = 49 * 2, where 49 is a perfect square, then we found \(\root{49} = 7\), resulting in \(\root{98} = 7\root{2}\). Simplifying each part makes subtraction or addition easier, as shown in the solution.
When simplifying square roots, look for perfect squares within the number. In the exercise, \(\root{72}\), we broke it down into \(36 \times 2\) because 36 is a perfect square. Then, we found \(\root{36} = 6\), so \(\root{72} = 6\root{2}\).
Similarly, for \(\root{98}\), we used that 98 = 49 * 2, where 49 is a perfect square, then we found \(\root{49} = 7\), resulting in \(\root{98} = 7\root{2}\). Simplifying each part makes subtraction or addition easier, as shown in the solution.
cube roots
Cube roots are about finding a number which, when multiplied by itself three times, gives the original number. For instance, the cube root of 27 is 3, because 3 * 3 * 3 = 27.
When dealing with cube roots in an exercise, look for perfect cubes within the number. For \(\root[3]{24}\), notice that 24 = 8 * 3, and 8 is a perfect cube. Thus, \(\root[3]{8} = 2\), leading to \(\root[3]{24} = 2\root[3]{3}\).
For \(\root[3]{81}\), 81 = 27 * 3, where 27 is a perfect cube, giving \(\root[3]{27} = 3\). This results in \(\root[3]{81} = 3\root[3]{3}\). Addition becomes straightforward as shown: \(\root[3]{24} + \root[3]{81} = 2\root[3]{3} + 3\root[3]{3} = 5\root[3]{3}\).
When dealing with cube roots in an exercise, look for perfect cubes within the number. For \(\root[3]{24}\), notice that 24 = 8 * 3, and 8 is a perfect cube. Thus, \(\root[3]{8} = 2\), leading to \(\root[3]{24} = 2\root[3]{3}\).
For \(\root[3]{81}\), 81 = 27 * 3, where 27 is a perfect cube, giving \(\root[3]{27} = 3\). This results in \(\root[3]{81} = 3\root[3]{3}\). Addition becomes straightforward as shown: \(\root[3]{24} + \root[3]{81} = 2\root[3]{3} + 3\root[3]{3} = 5\root[3]{3}\).
fourth roots
Fourth roots find a number which when used four times in a multiplication gives the original number. The fourth root of 16 is 2, because 2 * 2 * 2 * 2 = 16.
In simplifying fourth roots, look for perfect fourth powers. For \(\root[4]{80}\), knowing that 80 = 16 * 5, where 16 is a perfect fourth power, allows us to write \(\root[4]{16} = 2\) which simplifies to \(\root[4]{80} = 2 \root[4]{5}\).
Similarly, \(\root[4]{405}\) simplifies because 405 = 81 * 5, and \(\root[4]{81} = 3\). This results in \(\root[4]{405} = 3 \root[4]{5}\).
Calculations become simpler: \(\frac{1}{2}(\root[4]{80}) = \root[4]{5}\) and \(\frac{2}{3}(\root[4]{405}) = 2 \root[4]{5}\). The subtraction step is then \(\root[4]{5} - 2 \root[4]{5} = -\root[4]{5}\).
In simplifying fourth roots, look for perfect fourth powers. For \(\root[4]{80}\), knowing that 80 = 16 * 5, where 16 is a perfect fourth power, allows us to write \(\root[4]{16} = 2\) which simplifies to \(\root[4]{80} = 2 \root[4]{5}\).
Similarly, \(\root[4]{405}\) simplifies because 405 = 81 * 5, and \(\root[4]{81} = 3\). This results in \(\root[4]{405} = 3 \root[4]{5}\).
Calculations become simpler: \(\frac{1}{2}(\root[4]{80}) = \root[4]{5}\) and \(\frac{2}{3}(\root[4]{405}) = 2 \root[4]{5}\). The subtraction step is then \(\root[4]{5} - 2 \root[4]{5} = -\root[4]{5}\).
