in-3-38-write-each-expression-in-simplest-form-variables-in-the-radicand-with-an-even-index-are-no-7

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.
$$
\sqrt{250 a^{2}}+\sqrt{10 a^{2}}
$$

Julie Silva · Accepted Answer

we&#x27;re being asked to simplify the given expression. So remember, whenever you add radicals, you want to simplify all your radicals first. So let&#x27;s start with our first radical. We have the square root of 250 a square. Well, 250 is not a perfect square, however, the perfect square 25 devise into it 10 times. So I&#x27;m going to rewrite it as two square with a 25 times to square the 10 and then I&#x27;ll leave my a square than his own radical. Okay, so now let&#x27;s take a look at the second radical. We have the square of the 10 A squared Well, we can simplify the square, the 10 fervor because no perfect square goes into 10 evenly. However, a squared is a perfect square to square. The H word is a so we can simplify this toe a times to square the 10. All right, let&#x27;s go back to the first radicals and simplify well, the square, the 25 5, the square root of a square, this A and the square. The 10 can&#x27;t be simplified, so I&#x27;ll just bring that down. Then we&#x27;ll bring down our second radical, a Britain. So now both radicals air simplified and these two radicals are Expressions are right terms because both radic cans of the same. So we just need to combine the coefficients well. Five a plus a is six a. So our final answer will be six a Times Square the 10.

Julie Silva · Answer

for being asked to simplify the given expression. So we have the square with the two plus fiber to will remember, you can combine like radicals if they have the same radic. And in this case, because our Braddock an for both of our radicals are too we can combine these these air like radicals. So all we need to do is combine the coefficients. Remember, In the first radical to square, the two is coefficient is one So we&#x27;re gonna combine the coefficients. So we have one plus five, which is six. And then we bring down our radical the square the to. So our final answer is six route to

Julie Silva · Answer

we&#x27;re being asked to simplify the given expression. So we have the square of the 50 close to square the too well, they&#x27;re not like radicals. So what we need to do is simplify first. So let&#x27;s take a look at the square of the 50. Well, it&#x27;s not a perfect square, but the perfect square 25 goes into 52 times so we can rewrite this as the square of the 25 times the square the to. Now we can&#x27;t simplify the square with a two because no perfect square divides into it. So I&#x27;m just gonna bring it down. All right, let&#x27;s go ahead and simplify. Well, the square the 25 is five and the square this to can&#x27;t be simplified. So I&#x27;ll bring that down and then I&#x27;ll bring down our second radical, the square the to Now that both radicals air simplified, Now we can combine them because both rather cans of the same. So they combined them. We just need to combine the coefficients and remember the coefficient of the square. The two is one. Well, five plus one is six and then we&#x27;ll bring down our radical sign. So we have to square the to. So our final answer is six route to

Julie Silva · Answer

we&#x27;re being asked to simplify the given expression. So the first thing we need to do is simplify each of our radicals. So let&#x27;s start with our first term. We have a times to square the 45. So the perfect square nine goes into 45 9 times. So I&#x27;m gonna rewrite this as a times the square of the nine Times Square the five. Now, let&#x27;s look at our second radical. So we have the square of the 20 A squared. Well, the square. The 20 can be rewritten as the square of the four times five because the square root are sorry for is a perfect square. So we&#x27;re right. That&#x27;s a square. The four times square, the five and a square. Those are the perfect square. So I&#x27;m gonna leave that in its own radical Now for our last term, we can&#x27;t simplify the square with a two way, so we&#x27;re going to leave that as it&#x27;s all right. So let&#x27;s go back to the beginning. We&#x27;re gonna bring down a discover the nine is three. So I&#x27;m actually good. Put that in front of the A and you can&#x27;t simplify to square the five so We&#x27;ll just bring that down, bring down the plus sign. Then we have the square of the four, which is to, and the square with a squared, which is a and again we can simplify to square the five. So we&#x27;ll just bring that down. And as we mentioned earlier, we can&#x27;t simplify this laster. All right, great. All of our radicals air simplified. But I know this. Our first two terms are like terms because both of their radical ends of the same so we can combine them. So it&#x27;s add three. A Route five plus two, a Route five, which gives us five a Times Square, the five and we&#x27;re just gonna bring down that last term because it doesn&#x27;t have a like turn. And now we have simplified completely. So our final answer is five a Times Square, the five minus five times the square root of two a

Finian Lickona · Answer

So in this exercise, we are going to simplify the expression radical 287th minus a radical 58 of the seven. So the first thing I&#x27;m gonna do is I&#x27;m gonna try to simplify this radical as much as I can and then simplify this one to make it match up with this radical. So the first thing I notice is Aiken decomposed. This is this into two times 25 and I can rewrite this as a squared times a squared times he squared. Time&#x27;s a So if I factor all of the perfect square terms out, I get a five in the front and then I factored three ace squared terms outside becomes eight of the third. And what I&#x27;m left with under the radical is to a so I want to make this radical match up in the same way I noticed I can write the 200 as, uh, two times 100. And, of course, the same thing with this eight of the seventh term as was going on over here. So if I factor this 100 out of the radical, I get 100. My factor as many a terms I can like I did over here. Eight of the third and I am left with a to A under the radical. So now I simply group the new miracle coefficients 100 minus five times eight of the third radical to A and I am left with a final answer of 95 8 of the third radical to a

Julie Silva · Answer

we&#x27;re being asked to simplify to given expression. Remember, when you subtract radicals, you need to simplify both radicals first. Well, let&#x27;s take a look at the square root of eight A. Well, the perfect square four goes into 82 times. So I&#x27;m going to rewrite this as the square of the four times the square root of two way because four times to a is a day now I can&#x27;t simplify the square of the two way, so I&#x27;m gonna lead. That is this. Now let&#x27;s go back to our first term. We have the square with a four witches too. And we can&#x27;t simplify the square of the two way. So we bring that down. So we have two minus the square that you weigh minus the square that you weigh. All right, so now both radicals air simplified and we can combine these because both radic cans of the same. So we just combine the coefficients which remembered the coefficient of our second term is one. Well, two minus one is just one. So we&#x27;re just left with the square root of two a

Problem

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

Problem 9 Easy 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.
$$
\sqrt{250 a^{2}}+\sqrt{10 a^{2}}
$$

Answer

6$a \sqrt{10}$

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Ann Xavier Gantert

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6$a \sqrt{10}$

Algebra 2 and Trigonometry

Ann Xavier GantertAlgebra 2 and Trigonometry

Ann Xavier Gantert

Algebra 2 and Trigonometry