in-3-38-write-each-radical-in-simplest-radical-form-variables-in-the-radicand-of-an-even-index-ar-10

Question

In $3-38$ , write each radical in simplest radical form. Variables in the radicand of an even index are non-negative. Variables occurring in the denominator of a fraction are non-zero.
$$
\cdot \frac{1}{2} \sqrt{7} 2 a b^{5}
$$

Nick Johnson · Accepted Answer

Okay, so we have 1/2 time to be square with substitute times a times B to the fifth power. So to go ahead and try and simplify this well, we noticed that 72 is well, it&#x27;s a perfect square. 36 times to write 36 times two is 72. So you can go ahead and rewrite this as 1/2 times the square root of well, 36 times to which is 72. And then we still have a times with a B to the fifth. OK, Bose swear to 36 is six, so the six can come out. It was still left with the two under the radical. So the square 36 is six. They&#x27;re six times 1/2 or six times one after at 6/2, which is three. So this square 36 is six. That comes out of the radical and we&#x27;re left with again six times 1/2. That&#x27;s three and we still have the two under the square root. So we just have three times the square root of two and we still have a course is a be to the fifth, which you can&#x27;t do anything with so therefore, that&#x27;s as simple because we could go. You have three times the square to two times a be to the left.

Nick Johnson · Answer

we have the square root of a cube over B squared. Well, if we could go ahead if we could write something under the square room as the product of two things such that one of the things that perfect square. We didn&#x27;t pull out that perfect square. But do I really see any perfect squares here? Not really. Until unless right and we couldn&#x27;t write A cube is while a square times a so I could write this as the square root of a cube is a squared times a a cube right over the square root of five b. But that we have with a product of two things this world of, um, a score. Times A is the square root of a square times. The square root of eight with a square with a square is just a so this is equal to I&#x27;m a But then we still haven&#x27;t a under the square root, so I could write the square root of a cube could be rewritten. Must be a simpler or not. But as a times the square root of eight, okay, and we still have while the square root of five feet so over the square root of five B. So is that like, any simpler minutes? Well, I&#x27;m not sure about that, but I guess, um that&#x27;s about all we can do here. All right, take care.

Finian Lickona · Answer

So for this exercise, we&#x27;re going to simplify the term 2.42 radical to 0.42 a b to the fifth so we can rewrite to 0.42 as two times 1.21 you&#x27;ll recognize that this is just 121 divided by 100. And 100 and 21 is a perfect square, so we can rewrite this as 1.1 times 1.1, so we can fact this 1.1 outside of the radical. So we have 1.1 times the A term is gonna have to stay in the radical. But the beat of the fifth term could be rewritten as B squared times B squared, times B and those to be squares could be factored outs. That&#x27;s going to be be times be outside of the radical, so our final answer is going to be 1.1 B squared radical, a beat

Finian Lickona · Answer

So for this exercise, we&#x27;re going to be simplifying three over five beat of the third radical. Now three and five are both prime numbers, so they cannot be decomposed into any simpler terms and certainly not any perfect squares. So they&#x27;re going to have to stay inside the radical. Oh, I&#x27;m sorry I missed her at the problem. This should be a five. Let&#x27;s get the eraser tool and correct. All right, three beat fifth. Okay. So however, the beat of the fifth term can be re written as we&#x27;re just gonna put the 3/5 out in front. And now we have times one over B squared times one over B squared times one over B. And each of these b squared terms can be taken outside of the radical. So that&#x27;s gonna be one overby and one over Be So that&#x27;s gonna leave us with a one over B squared term outside of the radical. So our final answers one over B squared 3/5 times, one over B

Finian Lickona · Answer

All right, so we want to write the term eight Ah, b Cubed square root in its most simplified radical for So we&#x27;re gonna look for ways in which we can break each of these terms down into a product of some value and a perfect square. So if we look at the number eight, we recognize that we can write it as two times four. And if we look at the quantity, be cute. We recognize we can write it as b squared times be now, the square root of four is too. So we could take that out of the radical. Put it out here and this square root of B squared is be so we can also take that out of the radical and put it right here. So our final answer is going to be to be hang on to be. And this too has to stay in the radical because it can&#x27;t be simplified. And this be also has to stay in their radical. So our final answer is to be rude quantity to be

Finian Lickona · Answer

So for this exercise, we&#x27;re going to be simplifying the term route 15 over eight. Be cube. Now 15 cannot be decomposed into any perfect square, so it&#x27;s gonna have to stay inside the radical. However, the one over eight term can be rewritten, as is gonna include that 15 right there. One over two times four and four is a perfect square, so it can be taken outside of the radical. And then we can rewrite that beat of the third term as one over B squared times one over B. And of course, this one over B squared turn can be taken outside of the radical. So if we take that four outside of the radical becomes a 1/2. And if we take this B squared term outside of becomes one over B So our final answer is going to be 1/2 be radical. 15 over to be

Problem

In $3-38$ , write each radical in simplest radica…

Problem 15 Easy Difficulty

In $3-38$ , write each radical in simplest radical form. Variables in the radicand of an even index are non-negative. Variables occurring in the denominator of a fraction are non-zero.
$$
\cdot \frac{1}{2} \sqrt{7} 2 a b^{5}
$$

Answer

$3 \sqrt{2} a b^{5}$

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$3 \sqrt{2} a b^{5}$

Algebra 2 and Trigonometry

Ann Xavier GantertAlgebra 2 and Trigonometry

Ann Xavier Gantert

Algebra 2 and Trigonometry