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Mathematics 20 Online
OpenStudy (anonymous):

Add or subtract. Simplify by collecting like radical terms if possible. 3√162- √200+ 7√50 Answer was 52√2 Need an idea how they got there

OpenStudy (mathstudent55):

First, in order to simplify the square roots, you need to find the prime factors of 162, 200, and 50. Do you know how to find the prime factors of a number?

OpenStudy (anonymous):

9*18 for 162, 50*4 for 200, 25*2 for 50?

OpenStudy (mathstudent55):

Those products are correct, but they are not the full prime factorizations. Here they are: 162 = 2 * 3 * 3 * 3 * 3 200 = 2 * 2 * 2 * 5 * 5 50 = 2 * 5 * 5

OpenStudy (anonymous):

I see, what would i do next?

OpenStudy (anonymous):

Alright I get to that point so far

OpenStudy (mathstudent55):

Since we are dealing with square roots, we need factors that come in multiples of 2. \(3\sqrt{162} - \sqrt{200} + 7\sqrt{50}\) \(3 \sqrt{2 \times 3^4} - \sqrt{2^2 \times 2 \times 5^2 } + 7\sqrt{2 \times 5^2}\) \(3 \sqrt{2} \times \sqrt{3^4} - \sqrt{2^2} \times \sqrt{2} \times \sqrt{5^2 } + 7\sqrt{2} \times \sqrt{5^2}\) \(3 \sqrt{2} \times \color{red}{\sqrt{81}} - \color{red}{\sqrt{4}} \times \sqrt{2} \times \color{red}{\sqrt{25 }} + 7\sqrt{2} \times \color{red}{\sqrt{25}}\) Now you see which square roots can be simplified.

OpenStudy (mathstudent55):

Now we take the square roots of 25, 81, and 4.

OpenStudy (mathstudent55):

\(9 \times 3\sqrt{2} - 2 \times 5 \times \sqrt{2} + 5 \times 7\sqrt{2} \)

OpenStudy (mathstudent55):

\(27\sqrt{2} - 10 \times \sqrt{2} + 35\sqrt{2} \)

OpenStudy (mathstudent55):

Now we the terms: \(52 \sqrt{2} \)

OpenStudy (anonymous):

Ah I see thanks

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