Question
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Use the product rule to simplify the radical. \[ \sqrt{117} \]

Ask by Goodwin Fleming. in the United States
Jan 22,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

\[ \sqrt{117} = 3\sqrt{13} \]

Solution

To simplify the radical \(\sqrt{117}\) using the product rule, follow these steps: **Step 1: Factor the Number Inside the Radical** Find the prime factors of 117 to identify any perfect squares. \[ 117 = 9 \times 13 \] Here, 9 is a perfect square because \(9 = 3^2\). **Step 2: Apply the Product Rule for Radicals** The product rule states that \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\). Apply this to \(\sqrt{117}\): \[ \sqrt{117} = \sqrt{9 \times 13} = \sqrt{9} \times \sqrt{13} \] **Step 3: Simplify the Square Root of the Perfect Square** \[ \sqrt{9} = 3 \] **Step 4: Combine the Results** \[ \sqrt{117} = 3 \times \sqrt{13} = 3\sqrt{13} \] **Final Answer:** \[ \sqrt{117} = 3\sqrt{13} \]

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Beyond the Answer

Did you know that the number 117 has a fascinating prime factorization? It can be broken down as \( 3 \times 39 \) and then \( 39 \) can be further factored into \( 3 \times 13 \). This means \( 117 = 3^2 \times 13 \). Using the product rule for square roots, we can simplify \( \sqrt{117} \) into \( \sqrt{3^2 \times 13} \), which equals \( 3\sqrt{13} \). To grasp this concept better, think of square roots as pairs. Every time you find a perfect square within a radical, you can pull it out. With \( \sqrt{117} \), you recognized \( 3^2 \) as the perfect square, allowing you to simplify the expression joyfully, making \( \sqrt{117} \) much easier to work with. Happy simplifying!

Related Questions

Write down the next two terms for each geometric sequence. a \( 2 ; 4 ; 8 ; 16 ; \ldots \) b \( 5 ; 15 ; 45 ; \ldots \) c \( 3 ; 6 ; 12 ; 24 ; \ldots \) d \( 18 ; 6 ; 2 ; \ldots \) e \( 20 ; 10 ; 5 ; \ldots \) f \( 4 ;-12 ; 36 ; \ldots \) g \( 7 ;-14 ; 28 ; \ldots \) h \( 8 ; 4 ; 2 ; \ldots \) i \( \frac{1}{9} ; \frac{1}{3} ; 1 ; \ldots \) j \( 500(1,25) ; 500(1,25)^{2} ; 500(1,25)^{3} ; \ldots \) k \( 1000(1,8) ; 1000(1,8)^{2} ; 1000(1,8)^{3} ; \ldots \) i \( 6000(1,1) ; 6000(1,1)^{2} ; 6000(1,1)^{3} ; \ldots \) m \( 400\left(1+\frac{0,09}{12}\right) ; 400\left(1+\frac{0,09}{12}\right)^{2} ; 400\left(1+\frac{0,09}{12}\right)^{3} ; \ldots \) n \( 300\left(1+\frac{0,1125}{4}\right) ; 300\left(1+\frac{0,1125}{4}\right)^{2} ; 300\left(1+\frac{0,1125}{4}\right)^{3} ; \ldots \) o \( x\left(1+\frac{0,092}{2}\right) ; x\left(1+\frac{0,092}{2}\right)^{2} ; x\left(1+\frac{0,092}{2}\right)^{3} ; \ldots \) 2 Find the first three terms for each geometric sequence. a \( \mathrm{T}_{1}=2 \) and \( r=3 \) b \( \mathrm{T}_{1}=4 \) and \( r=\frac{1}{2} \) c \( \mathrm{T}_{1}=12 \) and \( r=\frac{-1}{3} \) d \( T_{1}=500 \) and \( r=1,1 \) e. \( \mathrm{T}_{1}=8000 \) and \( r=\left(1+\frac{0,09}{4}\right) \) f \( T_{1}=3 \) and \( T_{6}=96 \) g \( \quad T_{1}=7 \) and \( T_{5}=\frac{7}{81} \) h \( T_{2}=6 \) and \( T_{7}=192 \) \( T_{3}=18 \) and \( T_{5}=162 \) -d \( T_{3}=16 \) and \( T_{7}=256 \) k \( T_{2}=10 \) and \( T_{5}=80 \) I \( T_{2}=3 \) and \( T_{6}=\frac{1}{27} \) Determine: a which term is equal to 1280 in the sequence \( 5 ; 10 ; 20 ; \ldots \) b which term is equal to 1536 in the sequence \( 3 ; 6 ; 12 ; \ldots \) c which term is equal to 6561 in the sequence \( 3 ; 9 ; 27 ; \ldots \) d which term is equal to \( \frac{5}{16384} \) in the sequence \( \frac{5}{4} ; \frac{5}{8} ; \frac{5}{16} ; \ldots \) e which term is equal to \( \frac{1}{8192} \) in the sequence \( 16 ; 8 ; 4 ; \ldots \)
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