in-3-38-write-each-expression-in-simplest-form-variables-in-the-radicand-with-an-even-index-are-n-25

Question

In $3-38$ write each expression in simplest form. Variables in the radicand with an even index are non-negative. Variables occurring in the denominator of a fraction are non-zero.
$$
\sqrt{\frac{1}{6}}+\sqrt{\frac{8}{3}}-\sqrt{\frac{2}{3}}
$$

Julie Silva · Accepted Answer

Praveen has to simplify the given expression. So the first thing they do is rewrite each of our radicals using the quotient rule, meaning I&#x27;m gonna take the square root of 1/6 and rewrite it as the square of the one over the square. The six, then for our second term, will rewrite the square root of eight thirds as a square that they over the square, the three and for our last term will rewrite it as the square of the two over the square of the three. All right, let&#x27;s see if we can simplify any. It&#x27;s radicals when the first fraction we have to square with one, which is one. So that&#x27;s just going to be one over the square of the six. Now, for a second fraction, we have the square of the V. Well, the perfect square four goes into it two times and the square of the forest to so this can be rewritten as 2/2. So we&#x27;ll have to root two divided by the square of the three. And then our last radical can be simplified. So we just have the square of the 2/3. I&#x27;m sorry the square of the three. Don&#x27;t want to forget that. Now, remember when we added subtract fractions, we need a common denominator. So take a look at these last two. We can combine them because they have the same denominator and the numerator we have to route to minus route two, which is just a square the to. So we&#x27;re gonna bring down 1/6. Plus, we have to square with the to all over the square root of three. Now we want to go ahead and try and combine these fractions, which means we need a common denominator, which in this case, will be the square root of six. So I&#x27;m gonna multiply our second fraction by the square of the two over the square of the two. Now we have the square with the two times the square of the two, which is two square before, and the square of the three times to square. The to is to square the six and then we&#x27;ll bring down our first fraction. We&#x27;ll take a look. Your new Marie. There you have the square of the four, but that&#x27;s equal to two. So we&#x27;re going to have one over the square, the six plus two over the square of the six. Now we have a common denominator. Let&#x27;s go ahead and combine those new Meridor&#x27;s well. One plus two is three. So we&#x27;re gonna be left with three divided by the square, the six, and now we can&#x27;t simplify fervor. So our final answer, like I just said, is three divided by the square of the six.

Julie Silva · Answer

we&#x27;re being asked to simplify to given expression. So we have eight or three plus the square of the three. Well, these air considered to be like radicals because both radic ans are the same. So we just need to combine the coefficients now for that second radical. The coefficient of the square of the three is one. Remember, when you don&#x27;t see a coefficient, it&#x27;s always one. So now we&#x27;ll just combine our coefficients. So eight plus one is nine, and we&#x27;ll bring down our radical the square of the three, so our final answer is 93.

Finian Lickona · Answer

So in this exercise, we&#x27;re going to simplify the expression Route 24 plus to route three halves. Now I want to get these radical terms to match. So I&#x27;m going to have to find a way to make this 24 look like a three halves. I can rewrite this 24 as 48 over to and now the denominators of both radical terms match. So what I&#x27;m really going to do is find a way to decompose this 48 into some perfect squares. So I recognize that I can rewrite this as three times 16 over, too. And this 16 right here could be factored out as a four. So what I really have is four times three over route two, plus two times three over route to which, of course, simplifies to six. No times radical. Three over two

Finian Lickona · Answer

So in this exercise, we&#x27;re going to be simple. Find the expression before through of 48 minus the four through of three. Now three is not a perfect fourth, so we can simplify this radical any further. However we can rewrite 48 is 16 times three 16 is a perfect fourth. It&#x27;s two to the fourth power so we can bring it outside of the radical is it too? And we&#x27;re left with this three inside of the radical. And so now that these radicals are Oops, now that these radicals air matching, we can simply perform basic operations on them. So we have two minus one, which is simply just one fourth rule of three.

Julie Silva · Answer

we&#x27;re being asked to simplify the given expression. So the star, we&#x27;re gonna have to simplify each of these radicals. So let&#x27;s start with the square. The 12. Well, the perfect square four goes into it three times so we can rewrite this as to square the four Times square the three. Now, for our next radical, we have to square the 24 again. The perfect square four goes into it, but this time six times. So we write this as a square, the four times to square the six. All right for the square, the 48 the perfect square 16 goes into it three times. So we&#x27;ll rewrite this as to square the 16 times square the three. And for our last term, the perfect square nine goes into 27 3 times, so it will be rewritten as a square. The nine times square. The three. Now let&#x27;s go ahead and simplify well, the square of the forest to, and we can&#x27;t simplify to square the three, so we&#x27;ll bring that down. Next we have a square of the four again. It&#x27;s too, but you can&#x27;t simplify the square of the six. Next we have the square with 16 which is for And we can&#x27;t simplify the square of the three. And lastly, the square of the nineties three. And we can simplify the square of the three. So perfect. All of our radicals have now been simplified. Now we just have to combine any pairs of, like terms. Well, we have, like radicals that have radic cans of threes so we can combine all those coefficients. Well, two plus 46 and six plus three is nine. So that will give us nine. Route three and then we have minus two, Route six, which doesn&#x27;t have a late term. So we simply just bring that down. So now was simplified completely. So our final answer is nine or three minus two, Route six.

Finian Lickona · Answer

So in this exercise we are going to simplify the expressions. The cube root of two plus the cube root of 16. Now the Cuban too, can&#x27;t be simplified anymore. To is not a perfect you. However, we can rewrite 16 as two times eight and eight is a perfect cube. So if we take the cube root of eight outside of the radical we get to and we have this cube root left and there&#x27;s a two left inside the radicals. So we have radical curative two plus two cube Cuba too. So we just combine the coefficients their and our final answer is gonna be three radical cube root of to

Problem

In $3-38$ write each expression in simplest form.…

Problem 27 Hard Difficulty

In $3-38$ write each expression in simplest form. Variables in the radicand with an even index are non-negative. Variables occurring in the denominator of a fraction are non-zero.
$$
\sqrt{\frac{1}{6}}+\sqrt{\frac{8}{3}}-\sqrt{\frac{2}{3}}
$$

Answer

$\sqrt{\frac{3}{2}}$

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$\sqrt{\frac{3}{2}}$

Algebra 2 and Trigonometry

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Ann Xavier Gantert

Algebra 2 and Trigonometry