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In $3-38,$ solve each equation for the variable, check, and write the solution set.$$x+4 \sqrt{x}=5$$

$\{1\}$

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Chapter 3

REAL NUMBERS AND RADICALS

Section 8

Solving Radical Equations

Whole which of Numbers

Fractions and Mixed Numbers

Decimals

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In $3-38,$ solve each equa…

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02:28

03:01

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03:33

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four times the square root of X is equal to you. Fine. So well, let's go ahead. That subtracted the X over to just give us busy. Isolate our term with the square root of X in it. So we do that we get four times the square root of X is equal to well, negative X plus five. Either five minus X or negative X plus five. Okay, Now, if we square both sides here, we're gonna then eliminate the square root so four times the square root of X squared, that's gonna be you. Go to 16 X right? And then what? We square the right side. We have two terms here, so be careful. We have to square negative five plus X meaning we have negative negative experts. Five. That's negative. X plus five times itself. So native X plus five times negative ax plus five. OK, that's going to equal to or native X kind of data backs Get positive. X squared. Then we have native buybacks plus negative buybacks. That's negative. 10 acts and then plus five times five. So plus 25. So right. Negative X plus five squared is X squared minus 10 X plus 25. Therefore, we have 16 x is equal to X squared minus 10 X plus 25. Okay, let's go ahead and we have a quadratic. Now we can go ahead and subtract the 60 next term. So then set this quadratic equal to zero. So then let's see, we would get we have an X squared. We have my negative 10 X was attracting six x So negative 10 x minus 16 x That would be negative. 10 2026 x so minus 26 acts and then we have well plus 25 is well equal to zero. So now woken this factor. I believe it can write, um, cause this is going to be able to zero Now X squared has to be an x Times x 25 could be five times five right. But at the ad to a native 26 x, I have 25 times one right is 25 while 25 plus one after 26 in the same signs which in this case both have to be negative. And then we do add to a native 26 x for X squared minus one X minus 25 x is negatively six X and then minus 25 times one plus 25 equals zero. So we have two things being multiplied equal to zero. So at least one of those factors must be equal to zero. So therefore either X minus 25 is equal to zero, which would imply that X would be equal to 25 or X minus. One is equal to zero, which is implied that X would be equal to one. So then what? We have to go ahead. And it's a check that both of these solutions actually work. So we check X equals one. Well, we have one plus four times the square root of one. What a screwed of one is just one who would have one plus four times one that's one plus four. Right. We have one plus four being equal to five, which is true. We have five is he could've five that checks out. So therefore, X being equal toe one is a dot solution. How about ex being equal to 25 while we have 25 plus four times the square root of 25 being equal five so four times the square root of 25 would be four times five. So we would have five right? If x or not, or if I would have 25 x 25 would have 25 plus four times the square root of 25 4 times five, which is 20. We have 25 plus 20. Write those four times square with +55 is +54 times five is 2025 plus 20 would be equal to five. Is that true? No. Like I'll be 45 5 That is not true. So you can go ahead and can throw out X equal to 25. That is an extraneous solution. So our solution set is just a set containing the number one. So you write that as you set notation here. I was set braces, So we have a set containing the number one. So there is our solution sets. All right, take care

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