In Exercises 48–57, perform the indicated operations and write th...

Question

In Exercises 48–57, perform the indicated operations and write the result in standard form. √ (-32) - √ (-18)

Accepted Answer

Hello. Today we're gonna go ahead and solve a statement using imaginary numbers. So the given statement is the square root of negative 48 minus the square root of negative 27. Now I'm gonna go ahead and just rewrite the insides real quick. Rewriting the inside of the square roots is going to give me the square root of negative one times 48 minus the square root of negative one times 27. And if I go ahead and rewrite these into their own square roots, this is going to be the square root of negative one times the square root of 48 minus the square root of negative one times the square root of 27. Well, what can I do about this negative one square root here and here? Well, let's recall what the square root of negative one is equal to The square root of -1 is always going to result in an imaginary number. I so we can go ahead and rewrite this negative one. As just I so we're gonna have I times the square root of 48 minus I times the square root of 27. Now I can go ahead and factor out an eye from this statement here. So that's gonna give me I times the square root of - the square root of 27. Now, in order to simplify this, let's go ahead and rewrite 48 and 27 into a product of prime numbers. When we're looking at 48, could be broken up into two times two is a prime number. So I'm gonna go ahead and circle that. 24 is the same thing as 12 times two. So we have another instance of two here. 12 is six times 2 and six is going to be three times two. So rewriting this into a form of a square root Is going to be the square root of two times itself four times or two to the power of four times three. Since there's only one instance of three. Now we're gonna go ahead and do the same thing with 27. 27 is the same thing as saying nine times 33 is a prime numbers. I'll go ahead and circle it here and nine could be rewritten as three times three and both of these will be prime numbers. That means if I want to rewrite this, this is going to be the square root of three times itself three times or three cubed. So now what I have is going to be I times the square root of two to the power of four times three minus the square root of three. Cute. Now the square root allows me to pull out multiples of a number in groups of two. So looking at two to the power four, it's two times itself four times which results in two groups of two being pulled out. So I can go ahead and rewrite that as I times two squared times the square root of three. We can simplify the square root of three, so we'll go ahead and leave it in there and we're gonna go ahead and apply the same logic to here. We have three times itself three times mean that we can go ahead and pull out three once. So this is going to be rewritten as three times the square root of three. Now, just to work out what two squared is that's going to be I times four times the square root of three minus three times the square root of three. And now, since both of these quantities here have the square root of three, we can go ahead and subtract them from each other. So 4 -3 is just gonna be one or it's just gonna leave us with I times the square root of three. And that's gonna be our solution to this equation here. That means that the solution to this problem is going to be C. So I hope this helps you understand how to reduce square roots using imaginary numbers and I'll see you all in the next video

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