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In $3-38,$ solve each equation for the variable, check, and write the solution set.$$x=1+\sqrt{x+11}$$

$\{5\}$

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Chapter 3

REAL NUMBERS AND RADICALS

Section 8

Solving Radical Equations

Whole which of Numbers

Fractions and Mixed Numbers

Decimals

00:57

In $3-38,$ solve each equa…

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00:44

03:39

03:59

04:48

02:28

02:49

06:13

03:18

one plus the square root of X plus 11. So let's go ahead. That's first to break the ice like the square root. Let's subtract one from both sides of the equation, so we get that X minus. One is going to be equal to the square root of X plus 11. Okay, then, to get rid of the square, root the radical. Here. We can go ahead and square both sides of the equation, but be careful with squaring X minus one so X minus one squared means X minus one times X minus one X minus one times X minus one that X minus one squared that gives us an X squared and then in minus one x minus one x his A minus two X minus one times minus one plus one ex minus one square F minus one times X minus one is X squared minus two X plus one, and then that's gonna be equal to, well, what squaring both sides Well, that's going to be equal to the square root of X Plus 11 squared, which just gives us That's plus 11. So therefore we have X squared minus two X plus. One is equal to X plus 11. Okay. And now we can go ahead and get all our terms, toe one side of the equation and have a quadratic being set equal to zero. So the first go ahead and subtract X from both sides. The equation that gives us an X square and then negative two x minus X is gonna be negative. Three acts okay, and then was attracting 11 from both sides. Equation. So one minus 11 is gonna be well, I'm negative. 10 X squared, minus three X minus 10 is now equal to zero. So now we have a quadratic write in standard form that equal to zero. So can we go ahead and try and factor this quadratic? I think we can. So let's try and factor this. This is gonna factor. I believe as well. X squared half the x Times X now 10. That's five times two and the difference of five and two is three. Any different signs right to multiplied with negative 10 after add to a negative three extra. Also, the negative better beyond the bigger number or the five year X minus five times X plus two is equal to zero. So this would imply that X would be equal to five. Right? If X minus five is equal to zero, or or if X plus two is equal to zero, that would imply that X would be equal to negative two. Okay, let's go ahead and check now. The solutions to make sure they're actually solutions in our original equations, so X is equal. Five we would have five is equal 21 plus the square root of five plus 11. All right, well, five plus 11. That's 11 12 13 14 15 60 that 16. So we have one plus two squared. 16 square 16 is for So we had that. Five is equal toe one plus four, which is true, right, because five vehicle defiance So that checks out. So therefore exp indicative five is invalid solution. How about ex being equal to negative two xnegative to we would have negative to being equal 21 plus the square root of X plus 11. Now, before I even even try to even solve this. Wait a minute. The square root of X plus 11 has to be positive, right? What could be zero, but it can never cannot. The square root of a number cannot be negative. And then we're adding one and was somehow getting negative, too. So you can go ahead here and check out right and try toe put in negative, too. But this there's no way that one plus the square root of a number can be equal to a negative number. So this would not check out. And therefore, um, we can disregard negative two and just say that the solution set here is the sets containing the number five. All right, take a

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