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

$\{4,5\}$

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Chapter 3

REAL NUMBERS AND RADICALS

Section 8

Solving Radical Equations

Whole which of Numbers

Fractions and Mixed Numbers

Decimals

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In $3-38,$ solve each equa…

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the equation here, B minus three is equal to the square root of three B minus 11. We can go ahead and in square both sides the equation. So to get rid of this radical right, so we have B minus three squared or B minus three squared is B minus three times B minus three. So that's gonna be equal to be square. And then we have a minus three B minus three beats that's minus six B and then minus three times minus three is a positive nine. So be ministry squared is B squared minus six B plus nine that's gonna be equal to the square root of three B minus 11 squared. Which is just gonna give us what's under the radical. Give us just three B minus 11. All right, so now let's go ahead. And that's while move all our terms to one side of the equation would set a quadratic equal to zero. So it was attract the three B term. First, we're gonna subtract three big from both sides. Okay, so we still have a B squared on left side, and then we have on negative six B minus three beats. That's gonna be a negative nine b and then we're gonna add the 11. Right? We have three B minus 11. So when I add the 11 to both sides, we have 11 plus nine. What is that? That's while 10 plus nine is 19 plus the one more that's gonna be 20 b squared, minus nine B plus 20. And now this is gonna be equal to zero. So now we can go ahead and try and factor this quadratic we have here. So B squared minus nine B plus 20 is going toe factor as well. We have a B and B for B squared. 20 is gonna be while five times four right and five plus four is nine. So five and four should work here and we need opposite signs when he's running same size, same signs the ad to a on positive 20 but they haven't add to a negative nine. So both of these are gonna be negative. And there's just gonna be equal to zero. So therefore, if we have B minus five times, B minus four equals zero. The product of two things being equal to zero. At least one of the factors must be equal to zero. So either B minus five is equal to zero, which implies that B is equal to five or B minus four is equal to zero. That implies that B is equal to four. So either because you got a five or be is equal to four. Okay, let's go ahead and check both of these solutions and mix. Do they actually work in our original equation? So be is five. Well, then we have five minus three, which is gonna be equal to two. Right? And that's gonna be equal to the square root of well, three times times five minus 11. We have two is equal to the square root of five times because 15 of minus 11 is, uh, 432 is equal to describe it. A four which is true, right? Because it affords to we have to is equal to two that checks out. So five is a solution. How about four? Well, B is four. We have four minus three, which is equal to one, and then that's supposed to be equal to the square root of three B minus 11. So three times four times square root of three times four minus 11. So we have the square root of three times four minus 11 while that's becomes for his 12. The square root of 12 minus 11 which is one we have one, is equal to the square root of one, which is true, right? Because want this sort of one is one of one is equal to one. So that checks out. So we see that both solutions here five and four work out perfectly. Therefore, our solution set would be the set containing these two values, namely five and four. So there's a right that in set notation. That's the set containing five and four. All right, take care.

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